**Paragraph Type of Question**
**(A)** Electric potential at any point x, y, z in the space is given V = 4x

^{2} - 3x

**Question 1**
Find the electric field at any point (x, y, z)

a. (8x - 3)i

b. -(8x - 3)i

c. -8xi

d. 8xi

Solution
V = 4x^{2} - 3x

E_{x} = ∂V/∂x = 8x - 3

E_{y} = ∂V/∂y = 0

E_{z} = ∂V/∂z = 0

So, E = -(8x - 3)**i**

Hence (b) is correct

**Question 2**
Equipotential surface in the region are

a. planes parallel to X-Y plane

b. planes parallel to Y-Z plane

c. planes parallel to X-Z plane

d. none of the above

Solution
(b) is the correct choice

**Question 3**
Find the force experienced by a charge particle q at point (1, 1, 1)

a. -11qi

b. 11qi

c. 11qi + j

d. 11qi – j

Solution
E = -(8x - 3)**i**

F = qE = q (-8x - 3)**i **= q (-8 -3)**i **= -11q**i**

Hence (a) is correct

**Question 4**
Find the value of

a. -1 volt

b. 0 volt

c. -2 volt

d. 1 volt

Solution
$V_{(1,1,1)} – V_{(0,0,0)}= - \int_{0,0,0}^{1,1,1} \boldsymbol{E}.d\mathbf{L}$

$\int_{0,0,0}^{1,1,1} \boldsymbol{E}.d\mathbf{L}=V_{(0,0,0)}-V_{(1,1,1)} = -1 Volt$

Hence (a) is correct

**(B)** A spherical shell of radius R is carries a uniform surface charge q. Take the reference point at ∞

**Question 5**
Find the electric field at r>R .Here is the r is the unit vector across radial direction

Solution
From Gauss law

$\int \mathbf{E}.d\mathbf{S}$ = q_{in}/ε_{0}

E × 4πr^{2} = q/ε_{0}

$E =\frac {1}{4 \pi \epsilon _{0} } \frac {q}{r^2} $ radially outward

Hence (a) is correct

**Question 6**
Find the electric field at r<R

a. (1/4πε

_{0}) (q/R

^{2})

b. (1/4πε

_{0}) (q/r

^{2})

c. zero

d. (1/4πε

_{0}) (q/r

^{2} + R

^{2})

Solution
(c) is correct choice

**Question 7**
Find the potential at r>R

a. (1/4πε

_{0}) (q/r)

b. (1/4πε

_{0}) (q/r +R)

c. (1/4πε

_{0}) (q/r - R)

d. none of the above

Solution
$V = -\int_{\infty }^{r} E.dr$

= (1/4πε_{0}) (q/r)

Hence (a) is correct

**Question 8**
Find the potential at r<R

a. (1/4πε

_{0}) (q/r)

b. (1/4πε

_{0}) (q/R)

c. (1/4πε

_{0}) (q/R - r)

d. none of the above

Solution
Potential on the surface will be =(1/4πε_{0}) (q/R)

This will also be the potential inside

Hence (b) is correct

**Question 9**
If I placed an second uniformly charge shell (Q) at radius R'>R, will the value of potential change at r<R

a. increases

b. decreases

c. constant

d. none of the above

Solution
(a) is correct choice

**Question 10**
What will the electric field at r<R

a. (1/4πε

_{0}) (q/r

^{2})

b. (1/4πε

_{0}) (q/R

^{2})

c. zero

d. (1/4πε

_{0}) (q/r + R)

Solution
(c) is the correct choice

**Single Answer type question**
**Question 11**
A point charge q is located at (2, 3, 3) in xyz coordinate. Find the potential differences between A and B

A = (2, 3, 3)

B = (-2, 3, 3)

- q/ 4πε
_{0}
- 1/16πε
_{0}
- 3q/16πε
_{0}
- none of the above

Solution
V_{A} = kq/√[(x_{1} - x_{2})^{2} + (y_{1} - y_{2})^{2} + (z_{1} - z_{2})^{2}]

= kq/√(0 + 1^{2} + 0)

V_{B} =kq/√( 4^{2} + 0 + 0)

= kq/4

V_{A} - V_{B} = 3kq/4

= (3/4) × (1/4πε_{0}) q

= (3/16πε_{0}) q

Hence (c) is correct

**Question 12**
Two equally charges are placed on the x-axis at (-a, 0) and (a, 0). Charge is q and mass is m. They are released from rest. Find their velocities when they are 4a apart

a. q √(1/πε

_{0}m)

b. (q/4) √(1/πε

_{0}m)

c. q/πε

_{0}m

d. q/4πε

_{0}m

Solution
Law of conservation of energy

(U)_{start} - (U)_{end} = k_{end} - k_{start}

now k_{start} = 0

so, (q^{2}/4πε_{0} 2a) - (q^{2}/4πε_{0} 4a) = 2 × (1/2mv^{2})

(q^{2}/4πε_{0}) × (1/4a) = mv^{2}

v^{2} = q^{2}/16πε_{0}m

v = (q/4) √(1/πε_{0}m)

Hence (b) is correct

**Question 13**
A point charge q

_{1} = +2µC is placed at the origin of coordinates. A second charge

q

_{2} = -3µC is placed on the y-axis at y = 100cm. At what point on the y-axis, potential is zero

a. y = -200cm

b. y = 40cm

c. y = 200cm

d. none of the above

Solution
$V(y) =k[\frac {q_1}{|y-1|} + \frac {q_2}{|y|}]%

Setting V(y) = 0

$\frac {2}{|y|} - \frac {3}{|y-1|}]=0$

Or

2|y-1|=3|y|

we consider three cases to resolve the problem

y>1, 0<Y<1 and y < 0

Solving with this

y = 40cm y = -200cm

Hence (b) and (a) are correct

Class 12 Maths
Class 12 Physics