Tag: physics

Questions Related to physics

The electric potential V is given as a function of distance by $V=(5x^2+10x-4)volt$, where x is in metre. Value of electric field at $x=1m$ is :

physics coulomb's law field strength and potential gradient electric field as gradient of potential relation between electric field and electric potential

  1. $-23 V/m$

  2. $11 V/m$

  3. $6 V/m$

  4. $-20 V/m$

Show answer
Correct Option: D
Explanation:

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

or $\dfrac{dV}{dx}=10x+10$
The field, $E=-\dfrac{dV}{dx}=-(10x+10)$

At $x=1,  E=-(10+10)=-20 V/m$
So option D is correct. 

The potential at a point x (measured in $\mu m$) due to some charges situated on the x-axis is given by $V(x)=20/(x^2-4)volt$
The electric field E at $x=4\mu m$ is given by :

physics coulomb's law field strength and potential gradient electric field as gradient of potential relation between electric field and electric potential

  1. $(10/9)volt /\mu m$ and in the $+ve$ x direction

  2. $(5/3)volt /\mu m$ and in the $-ve$ x direction

  3. $(5/3)volt /\mu m$ and in the $+ve$ x direction

  4. $(10/9)volt /\mu m$ and in the $-ve$ x direction

Show answer
Correct Option: A
Explanation:

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.

A and B are two points in an electric field. If the work done in carrying $4.0 C$ of electric charge from A to B is $16.0 J$, the potential difference between A and B is :

physics coulomb's law field strength and potential gradient electric field as gradient of potential relation between electric field and electric potential

  1. $zero$

  2. $2.0 V$

  3. $4.0 V$

  4. $16.0 V$

Show answer
Correct Option: C
Explanation:

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

Determine the electric field strength vector if the potential of the field depends on x, y coordinates as $V = a (x^2 - y^2)$, where a is a constant.

physics coulomb's law field strength and potential gradient electric field as gradient of potential relation between electric field and electric potential

  1. $\vec{E} = - 2a(x\widehat{i} - y\widehat{j})$

  2. $\vec{E} = - a(x\widehat{i} - y\widehat{j})$

  3. $\vec{E} = - \dfrac{a(x\widehat{i} - y\widehat{j})}{2}$

  4. $\vec{E} = - \dfrac{a(x\widehat{i} - y\widehat{j})}{4}$

Show answer
Correct Option: A
Explanation:

$ \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 )  $

Determine the electric field strength vector if the potential of the field depends on x, y coordinates as $V = axy$ , where $a$ is a constant.

physics coulomb's law field strength and potential gradient electric field as gradient of potential relation between electric field and electric potential

  1. $\vec{E} = -a(y\widehat{i} + y\widehat{j})$

  2. $\vec{E} = -a(x\widehat{i} + y\widehat{j})$

  3. $\vec{E} = -a(x\widehat{i} + x\widehat{j})$

  4. $\vec{E} = -a(y\widehat{i} + x\widehat{j})$

Show answer
Correct Option: D
Explanation:

$ \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 ) $

The electric potential existing in space is $V(x, y, z) = A (xy+ yz + zx)$. Find the expression for the electric field :

physics coulomb's law field strength and potential gradient electric field as gradient of potential relation between electric field and electric potential

  1. $-A{(x + Z) \widehat{i} + (y + Z) \widehat{j} + (x + y) \widehat{k}}$

  2. $-A{(y + Z) \widehat{i} + (x + Z) \widehat{j} + (x + y) \widehat{k}}$

  3. $-Ax{ \widehat{i} +y \widehat{j} + Z\widehat{k}}$

  4. $-A{(x+y) \widehat{i} + (x + y) \widehat{j} + (x + y-2Z) \widehat{k}}$

Show answer
Correct Option: B
Explanation:

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

At a certain distance from a point charge, the field intensity is 500 V/m and the potential is 3000 V. The distance and the magnitude of the charge respectively are :

physics coulomb's law field strength and potential gradient electric field as gradient of potential relation between electric field and electric potential

  1. 6 m and 6 $\mu $C

  2. 4 m and 2 $\mu$C

  3. 6 m and 4 $\mu$C

  4. 6 m and 2 $\mu$C

Show answer
Correct Option: D
Explanation:

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

In a certain region of space, the electric potential is $V (x, y, z) = Axy - Bx^2$ $+Cy$, where $A, B\ and\ C$ are positive constants. Calculate the $x, y\ and\ z$ components of the electric field.

physics coulomb's law field strength and potential gradient electric field as gradient of potential relation between electric field and electric potential

  1. $E _x = - Ax, E _y = -Ay, E _z = 0$

  2. $E _x = - Ax + 2Bx, E _y = -Ay -C, E _z = 0$

  3. $E _x = - Ay + 2Bx, E _y = -Ax -C, E _z = 0$

  4. $E _x = - Ay, E _y = -Ax, E _z = 0$

Show answer
Correct Option: C
Explanation:

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


So, $E _x=-\dfrac{V}{dx}=-[Ay-2Bx]=2Bx-Ay$;


$E _y=-\dfrac{dV}{dy}=-[Ax+C]=-Ax-C$ and 

$E _z=-\dfrac{dV}{dz}=0$

Potential difference between centre and surface of the sphere of radius R and uniform volume charge density $\rho$ within it will be :

physics coulomb's law field strength and potential gradient electric field as gradient of potential relation between electric field and electric potential

  1. $\displaystyle \dfrac{\rho R^2}{6 \varepsilon _0}$

  2. $\displaystyle \dfrac{\rho R^2}{4 \varepsilon _0}$

  3. $\displaystyle \dfrac{\rho R^2}{3 \varepsilon _0}$

  4. $\displaystyle \dfrac{\rho R^2}{2 \varepsilon _0}$

Show answer
Correct Option: A
Explanation:

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 $V=-\int _R^0 E.dr=-\int _R^0 \dfrac{\rho r}{3\epsilon _0} dr=\dfrac{\rho R^2}{6\epsilon _0}$

A uniform electric field exists in x-y plane. The potential of points A (-2m, 2m), B(+2m, 2m) and C(2m, 4m) are 4 V, 16V and 12 V respectively. The electric field is :

physics coulomb's law field strength and potential gradient electric field as gradient of potential relation between electric field and electric potential

  1. $(4\widehat{i} + 5 \widehat{j}) V/m$

  2. $(3\widehat{i} + 4 \widehat{j}) V/m$

  3. $-(3\widehat{i} + 4 \widehat{j}) V/m$

  4. $(3\widehat{i} - 4 \widehat{j}) V/m$

Show answer
Correct Option: D
Explanation:
Let equation of potential be $ax+by+c$ where $(x,y)$ are the co-ordinates of the point in $x, y$ plane.
so, $a(2)+b(2)+c=4$
$a(-2)+b(2)+c=16$
$a(2)+b(4)+c=12$
Solving above equations, we get $a=-3;b=4;c=2$
so equation of potential is $V=-3x+4y+2$ 
Now the electric field is $\vec E=(-dV/dx)\vec i+(-dV/dy)\vec j=3\vec i-4\vec j$