Electric Field and Electric Potential Relationship

Electric field lines point from higher to lower potential. Moving in the direction opposite to the lines of force means moving toward higher potential.

$E = \dfrac{V}{r} = \dfrac{10V}{10 \times 10^{-2}m} = \dfrac{10 \times 100}{10} = 100 \dfrac{v}{m}$

Electric field lines are defined as the path along which a positive test charge would move. Since a positive charge moves from higher to lower potential, the potential decreases along the lines of force.

The negative sign is just a convention and it signifies that the direction of E is opposite to the direction in which potential increases.

The electric field is the negative gradient of potential: E = -∇V. Computing partial derivatives: ∂V/∂x = -2xy√z, ∂V/∂y = -x²√z, ∂V/∂z = -x²y/(2√z). At point (2,1,1): Ex = -2(2)(1)(1) = -4, Ey = -(4)(1) = -4, Ez = -(4)(1)/(2×1) = -2. Therefore E = -(-4i - 4j - 2k) = 4i + 4j + 2k. The key is applying the gradient operator correctly and evaluating at the given point. Note that option D has a typo (should be î, not 1̂) but is clearly the intended answer.

In a shell $E=0 $ but $V \neq 0$.
Along the equatorial line of a dipole, $V=0 $ but $E  \neq 0$.

given force$ = 2.5$ $\mu$
we know $F = Eq$
$\Rightarrow 2.5=E( 6.25\ \mu c)$ 

The field is uniform. Hence ðv/ðr=constant=p(let)
Hence v=V°+p(i+j+k)
Now at origin v=V°=10volt
Also at the 3 points the value of the voltage are 8volt each. Hence p= -2
At (1,1,1) The voltage is=10 -2(1+1+1)
                                 =4 volt(ans)

E = -dV/dr. Given E = 10, V = -integral(E dr). Potential difference V(0,0) - V(10,20) = integral from 0 to 10 of E dx + integral from 0 to 20 of E dy. This requires more context on the field vector. Assuming a uniform field, the result is 500.

$V=6x-8xy^2$

E = -dV/dx. The potential changes by 40 V over 4 m. The average field is 10 V/m. Since it decreases uniformly, the field is constant at 10 V/m everywhere in that interval.

B. $3i+2j+6k$

$\quad E∝r\quad and\quad V=0\quad at\quad r=0$

Given, $V=k[2x^2-y^2+z^2]$

Electric field , $\vec{E}=-(\dfrac{dV}{dx}\hat i+\dfrac{dV}{dy}\hat j+\dfrac{dV}{dz}\hat {k})$

or,$\vec{E}=-k(4x\hat i-2y\hat j+2z\hat k)$

or,$\vec{E} _{(1,1,1)}=-k(4\hat i-2\hat j+2\hat k)$

Magnitude of electric field$ =|\vec{E} _{(1,1,1)}|=\sqrt{k^2(16+4+4)}=k\sqrt {24}=2k\sqrt 6$

Electric field and electric potential are related by E = -grad(V). A region can have zero electric field while having a non-zero constant electric potential (such as inside a charged conducting sphere). Thus, E can be zero where V is non-zero.

The electric field E between two parallel plates is given by E = V/d. Here, V = 10 V and d = 1 cm = 0.01 m, so E = 10 / 0.01 = 1000 N/C.

Now equation of equipotential surface is $y=2x$
Now electric field along the euipotential surface should be zero
therefore angle made by equipotential surface with x-axis is $tan^{-1} { (2) } $
Now since net electric field should be perpendicular to the equipotential surface
therefore for any electric field which makes an angle $tan^{-1} { (-1/2) } $ with x-axis can be the electric field at point $(1,2)$ which is true only for option (D)

Given,

$V=2x+3y-z$

$E _x=-\dfrac{dV}{dx}=-2,  E _y=-\dfrac{dV}{dy}=-3 $ and $E _z=-\dfrac{dV}{dz}=1$

As the field components are independent of x,y and z so the field remains constant.

uniform electric field means the electric field vector does not vary with positive and electric field lines are parallel and equally spaced. the statement given define the equipotential surface, so answer is (d) -none of these.

The electric field is $E=-\dfrac{dV}{dx}$
If $V=0$, we can not say $E$ must be zero, we say only $E$ may be zero.
If $V \neq 0 $, $E$ must be zero when $V$ is max i.e, $\dfrac{dV}{dx}=0$ For example, inside the conductor $E=0,$ but $V \neq 0 $
If $E \neq 0$ , $V$  may be zero when two equal and opposite charges separated by a distance and  at the midpoint in between the charges field is non-zero but potential is zero.

$F=qE \Rightarrow E=\dfrac{F}{q}=\dfrac{2.5}{6.76\times 10^{-6}}=3.7\times 10^5$

The potential gradient is the electric field i.e, $E=-\nabla V=-3.7 \times 10^5 $

As the electric potential decreases uniformly so the electric field is uniform along the x axis.

$E=-\dfrac{dV}{dx}=\dfrac{120-80}{1-(-1)}=20 $V/m. It must be equal to 20 V/m at the origin.

Electric field is directly propotional to $r$, therefore
$E=kr$
We know that
$\ V=-\int _{ 0 }^{ r }{ \overrightarrow { E. } \overrightarrow { dr }  } $
which gives 
$V=-\dfrac { k{ r }^{ 2 } }{ 2 } $ since V at $r=0$ is $0$
Hence V is proportional to ${ r }^{ 2 }$

Here, $\vec{E}=20 \hat i \Rightarrow E _x=20$
Potential difference ,$V _B-V _A=-\int _A^B Edr=-\int _4^6E _x dx=-20\int _4^6 dx=-20(6-4)=-40  V$

The relation between electric field and potential is given by $\vec{E} = \displaystyle -\frac{\partial V}{\partial x}\hat{i} -\frac{\partial V}{\partial y}\hat{j} $

For a charged conducting sphere, the electric field at the surface is E = kQ/R^2 and the potential is V = kQ/R. Therefore, V = E * R. Given E = 3 * 10^6 V/m and R = 1 m, V = 3 * 10^6 V.

Work done by the electric field is W = q * (V_A - V_B). For an electron, q = -1.6 * 10^-19 C. Given W = 6.4 * 10^-19 J, we have 6.4 * 10^-19 = -1.6 * 10^-19 * (V_A - V_B), which simplifies to V_A - V_B = -4 V, or V_B - V_A = 4 V.

The electric field vector is always perpendicular to the equipotential line at any given point. The slope of the equipotential line y = 2x is m1 = 2. Therefore, the slope of the electric field vector must be the negative reciprocal, m2 = -1/2. Testing the options for a vector (E_x, E_y) with slope -1/2 = E_y / E_x: option D gives E_x = -8 and E_y = 4, so E_y / E_x = 4 / (-8) = -1/2, which matches correctly.

$V(x)=5-4x^2$

$V(1)=5-4=1 V,  V(2)=5-4(2^2)=-11 V$

Potential difference between $x=1 m$ and $x=2 m$ is $V _{12}=V _1-V _2=1-(-11)=12 V$

here, $E _x=-\dfrac{dV}{dx}=8x,  E _y=-\dfrac{dV}{dy}=0$ and $E _z=-\dfrac{dV}{dz}=0$

The electric force on $1$ coulomb charge at $x=1$ is $F=qE _x=1(8)=8 N$

As the field E is uniform, so E is constant at every point.
As E is directed parallel to x axis, so $\vec{E}=E\hat i$
The work done , $W=\int \vec{F}.\vec{dr}=\int qE\hat i.(\hat{i}dx+\hat{j}dy+\hat{k}dz)$
$W=qE\int dx=qE[\int _a^{2a}dx+\int^a _{2a}dx+\int _a^{0}dx]=qE[2a-a+a-2a+0-a]=-qEa$

The electric field is E = -grad(V). E = -(dV/dx i + dV/dy j + dV/dz k). Given V = k(2x^2 - y^2 + z^2), E = -k(4x i - 2y j + 2z k). At (1, 1, 1), E = -k(4 i - 2 j + 2 k). The magnitude is k * sqrt(4^2 + (-2)^2 + 2^2) = k * sqrt(16 + 4 + 4) = k * sqrt(24) = 2k * sqrt(6).

The electric field E = F/q = 3000 N / 3 C = 1000 N/C. The potential difference V = E * d. With d = 1 cm = 0.01 m, V = 1000 * 0.01 = 10 V.

The electric field in the x-direction is given by E = -dV/dx. Given V(x) = 20 / (x^2 - 4), differentiating with respect to x using the quotient rule gives dV/dx = 20 * (-1) * (2x) / (x^2 - 4)^2 = -40x / (x^2 - 4)^2. At x = 4, dV/dx = -40(4) / (16 - 4)^2 = -160 / (12^2) = -160 / 144 = -10/9. Since E = -dV/dx, E = -(-10/9) = +10/9 Volt/um, and the positive sign indicates it is in the positive x direction.

$dV=-\vec{E}.d\vec{r}=-Edr cos\theta$

Potential decreases in the direction of electric field. So it depends  on whether the lines of forces are from $A$ to $B$ or from $B$ to $A$.

$\displaystyle F = qE$
work will only be done in moving the charged particle in $x$ direction

Work done in the presence of electric field E is $W=\vec F. \vec r = q\vec E.\vec r$
$W=Q[(E _1\hat{i}+E _2\hat{j}).( a\hat{i}+b\hat{j})]$
$W=Q(E _1a+E _2b)$

$dv=-\vec E.\vec dx=-E dx \cos\theta$

$\displaystyle E = -\dfrac{dV}{\cos\theta dx} $

$\cos\theta\approx 1$, if we take $\cos\theta =1$,then

$E _{min}= -\dfrac{80-120}{1-(-1)} =\dfrac{40}{2}= 20 V cm^{-1}$

Hence E would be greater than or equal to $20$

Since $E=\dfrac{-dV}{dr}$

A zero potential at a point never means that the electric field is also zero at a point always.
$E=0$ when $V=$ constant.
If $V=0$; then $E$ may or may not be zero.
Similarly, if $E=0$; then $V$ must either be a constant or may be zero.
so correct option is(d).

For a uniform field E = E0, V = -integral(E dx) = -E0 * x + C. Since V(0) = 0, C = 0. Thus, V(x) = -E0 * x.

$\displaystyle E _x = -\dfrac{dV}{dx} = -\dfrac{1}{2} [-4] = 2$

 $\displaystyle E _y = -\dfrac{dV}{dy} = -\dfrac{1}{2}[2y] = - y = -1$

 $\displaystyle \therefore \vec {E} = E _x \hat{i}+E _y \hat{j}=2\hat{i}-1\hat{j}$

E = -dV/dx. V = 4x^2, so E = -d(4x^2)/dx = -8x. At x = 1, E = -8 V/m. The negative sign indicates the direction is along the negative x-axis.

$dV=-\vec E.\vec dx-\vec E.\vec dy$
$\Delta V=-\int Edx-\int Edy$
$\displaystyle V _B-V _A = -(\int _{1}^{2}2dx+\int _{2}^{1}3dy)$

$\displaystyle =-[2(2-1)+3(1-2)]$
$\displaystyle =-[2-3] = 1 V$
 Hence, $V _A-V _B = -1 V$

The electric field of an infinite sheet is E = sigma / (2 * epsilon_0). Given sigma = 10^-7 C/m^2 and epsilon_0 = 8.85 * 10^-12 F/m, E = 10^-7 / (2 * 8.85 * 10^-12) = 10^5 / 17.7 = 5649 V/m. Using V = E * d, d = V / E = 5 / 5649 = 0.000885 m = 0.88 mm.

Given, $V=5x^2+10x-4$

Given, $\displaystyle V(x)=\dfrac{20}{x^2-4}$

Electric field , $\displaystyle E=-\dfrac{dV}{dx}=-\dfrac{-20}{(x^2-4)^2}(2x)=\dfrac{40x}{(x^2-4)^2}$

At $\displaystyle x= 4 \mu m,   E=\dfrac{40(4)}{(4^2-4)^2}=\dfrac{160}{144}=(10/9)  volt/\mu m$

Positive sign indicates that $\vec{E}$ is in the +ve x-direction.

The work done, $W _{A\rightarrow B}=q\int _{V _A}^{V _B}dV=q(V _B-V _A)$
or $16=4(V _B-V _A) \Rightarrow V _B-V _A=4  V$

$ \vec E = - \triangledown V $   ( negative of gradient of V)

$\vec E = - (\dfrac{dV}{dx} \hat i + \dfrac{dV}{dy} \hat j )=-a(2x \hat i +2y \hat j )  $

$ \vec E = - \triangledown V $   ( negative of gradient of V)

$\vec E = - (\dfrac{dV}{dx} \hat i + \dfrac{dV}{dy} \hat j )=-a(y \hat i + x \hat j ) $

$ \vec E = - \triangledown V = -A[(y+z) \hat i + ( z+x) \hat j +( y+x) \hat k ] V/m $

The electric field at distance d due to point charge q is $E=kq/d^2 $ and potential $V=kq/d$ 

so, $E=V/d $ or $ d=\dfrac{V}{E}=\dfrac{3000}{500}=6 m$

since, $V=kq/d $

or $3000=9\times 10^9\times \dfrac{q}{6} $

or $q=2\times 10^{-6} C=2 \mu C$

Here, $V(x,y,z)=Axy-Bx^2+Cy$

Using Gauss's law the electric field inside the sphere , $E.4\pi r^2=\dfrac{q}{\epsilon _0}=\rho\dfrac{(4/3)\pi r^3}{\epsilon _0}$

or $E=\dfrac{\rho r}{3\epsilon _0}$

Potential difference between surface and center is $V=-\int _R^0 E.dr=-\int _R^0 \dfrac{\rho r}{3\epsilon _0} dr=\dfrac{\rho R^2}{6\epsilon _0}$

$ \vec E = - \triangledown V = - [ ( Ay - 2Bx) \hat i + ( Ax + C) \hat j ] $

$ \vec E = - \triangledown V = -A[(y+z) \hat i + ( z+x) \hat j +( y+x) \hat k ] = -10[2 \hat i + 2 \hat j + 2 \hat k ] = 20\sqrt{3} N/C$

If electric field at some point is zero, it is not necessary for the electric potential to be the same. Consider the mid point of line joining to equal charges of same sign. Field there is zero, but potential is finite and positive.

We know that , $F=-\Delta U$, change in potential energy. 

We know that, $E=-\dfrac{dV}{dx}$

$dV=-E.dr$, so the electric potential $V$ decreases continuously as we move along the direction of the electric field

given potential $=x{ y }^{ 2 }{ z }^{ 2 }$

$4\times 10^{20}eV=4\times 10^{20}\times 1.6\times 10^{-19}=64 J$

In the direction of electric field the electric potential decreases. This is because electric potential is the work done against the direction of electric field.

Potential difference between two points is given by    $\Delta V = -E.r$
Given :  $E = 0$
$\implies  \ \Delta V = 0$
$\implies \ V =$ constant

The electric potential decreases uniformly from V to -V over a distance of 2x_0 along the x-axis, meaning the magnitude of the uniform electric field is the total potential difference divided by the total distance. The total change in potential is 2V over a distance of 2x_0, giving an electric field magnitude of V/x_0. Since the question asks for the field at the origin and specifies uniform decrease, it must equal this value.

The electric field intensity is the negative gradient of the potential V. Taking partial derivatives, Ex = -dV/dx = 5, Ey = -dV/dy = 5, and Ez = -dV/dz = 5. The magnitude of the electric field is the square root of the sum of the squares of these components, which is sqrt(5^2 + 5^2 + 5^2) = sqrt(75) = 5*sqrt(3) V/m.

From photo electric effect equation :

Now , Since both are infinite plates and carry same charge density therefore, electric field between them will be equal to zero.
Now, potential will be constant between them and at $x=0; V=0$ and V = constant between the plates.
Therefore, V=0 between the plates means $0\le x\le a$
Now electric field beyond $x=a$ is $2\times \sigma/2\epsilon _o=\sigma/\epsilon _o$
We know that,
$V=-\int _{ a }^{ x }{ \overrightarrow { E } .\overrightarrow { dx }  } $
$V=-E(x-a)$
$V=-\sigma(x-a)/\epsilon _o$

and for $x<0$
$V=-\int _{ x }^{ 0 }{ \overrightarrow { E } .\overrightarrow { dx }  } $
$V=Ex$
$V=\sigma x/\epsilon _o$
option (A)(B)(D) are correct.

Work done , $W=\int \vec{F}.\vec{dr}$

$ \overrightarrow{E} = -\triangledown \phi $
$ \overrightarrow{\triangledown}.\overrightarrow{E} = \rho/\epsilon _0 $
Now, $\phi = \phi _0 (x^2 + y^2 + z^2) \Rightarrow \overrightarrow{E} = -2\phi _0 ( \hat{i}+\hat{j}+\hat{k} ) \Rightarrow \rho = -6\phi _0 \epsilon _0 $

Given, $\vec{E}=x\hat{i}+2y\hat{j}+3\hat{k}$ and $ q=2 C$
Work done, $W=\int^B _Aq\vec{E}.\vec{dr}=q\int^B _A(x\hat{i}+2y\hat{j}+3\hat{k}).(dx\hat{i}+dy\hat{j}+dz\hat{k})$
or,$W=2\int^{(4,2,1)} _{(3,3,1)}xdx+2ydy+3dz=2[\frac{16-9}{2}+(4-9)+3(1-1)]=7-10=-3$
Magnitude of work done$=|W|=3$

Potential gradient is nothing but the rate of change of electric potential with position and it is equal to electric field at that point.

$\vec{E}=-\dfrac{\partial V}{dx}\hat{i}-\dfrac{\partial V}{dy}\hat {j}-\dfrac{\partial V}{dz}\hat{k}$

$U = \dfrac{{k,Q \cdot Q}}{{2R}}$

$ E _1 $ = Electrical field due to $+Q$
$E _2$ = Electrical feild due to $-2Q$
There resultant is $0$ at this point