Tag: physics

Quizzes (598) Questions (13441)

Questions Related to physics

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

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 :

  1. $-23 V/m$

  2. $11 V/m$

  3. $6 V/m$

  4. $-20 V/m$
Reveal answer Fill a bubble to check yourself
D Correct answer
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. 

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

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 :

  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
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

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

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 :

  1. $zero$

  2. $2.0 V$

  3. $4.0 V$

  4. $16.0 V$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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

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

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.

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

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

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

  4. <span>$\vec{E} = - \dfrac{a(x\widehat{i} - y\widehat{j})}{4}$</span>
Reveal answer Fill a bubble to check yourself
A Correct answer
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 )  $

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

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.

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

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

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

  4. <span>$\vec{E} = -a(y\widehat{i} + x\widehat{j})$</span>
Reveal answer Fill a bubble to check yourself
D Correct answer
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 ) $

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

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

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

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

  3. <span>$-Ax{ \widehat{i} +y \widehat{j} + Z\widehat{k}}$</span>

  4. <span>$-A{(x+y) \widehat{i} + (x + y) \widehat{j} + (x + y-2Z) \widehat{k}}$</span>
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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

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

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 :

  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
Reveal answer Fill a bubble to check yourself
D Correct answer
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 $q=2\times 10^{-6} C=2 \mu C$

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

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.

  1. <span>$E _x = - Ax, E _y = -Ay, E _z = 0$</span>

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

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

  4. <span>$E _x = - Ay, E _y = -Ax, E _z = 0$</span>
Reveal answer Fill a bubble to check yourself
C Correct answer
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 

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

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

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

  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. $\displaystyle \dfrac{\rho R^2}{2 \varepsilon _0}$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $V=-\int _R^0 E.dr=-\int _R^0 \dfrac{\rho r}{3\epsilon _0} dr=\dfrac{\rho R^2}{6\epsilon _0}$

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

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 :

  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. $(3\widehat{i} - 4 \widehat{j}) V/m$
Reveal answer Fill a bubble to check yourself
D Correct answer
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 $\vec E=(-dV/dx)\vec i+(-dV/dy)\vec j=3\vec i-4\vec j$