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NCERT Class XII Chapter
Electrostatic Potential and Capacitance
Questions With Solutions

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Question : 14
Total: 37
Two tiny spheres carrying charges 1.5 μC and 2.5 μC are located 30 cm apart. Find the potential and electric field:
(a) at the mid-point of the line joining the two charges, and
(b) at a point 10 cm from this midpoint in a plane normal to the line and passing through the mid-point.
Solution:  
(a) At mid point C of line joining two charges, electric potential is
VC = CCA+VCB
or VC =
1
4πε0
 [
qA
rCA
+
qB
rCB
]
= 9 × 109[
1.5
15
+
2.5
15
] ×
10−16
10−2
or VC = 2.4 × 105 V
and electric field at point C is
EC = ECB−ECA [As ECB > ECA and they are directed opposite to each other]
or EC =
1
4πε0
 [
qB
rCB2
+
qA
rCA2
] = 9 × 109[
2.5
152
−
1.5
152
] ×
10−6C
10−4m2
or EC = 4 × 105Vm−1
directed in direction of
→
ECB
 i.e. from C to A.
(b) At given point P on perpendicular bisector of line joining two charges, electric potential is
VP = VPA+VPB or VP =
1
4πε0
[
qA
rPA
+
qB
rPB
] = 9 × 109[
1.5
18
+
2.5
18
] ×
10−6C
10−2m
or VP = 2.0 × 105 V
and horizontal component of net electric field at point P is
Ex = EPA cos θ - EPB cos θ = (EPA−EPB) cos θ
or Ex =
1
4πε0
[
qA
rPA2
+
qB
rPB2
] cos θ = 9 × 109[
1.5
182
2.5
182
] ×
10−6
10−4
×
15
18
or Ex = - 2.3 × 105Vm−1
whereas vertical component of net electric field at point p is
Ey = EPA sin θ + EPB sin θ = [EPA+EPB] sin θ
or Ey =
1
4πε0
[
qA
rPA2
+
qB
rPB2
] sin θ = 9 × 109[
1.5
182
+
2.5
182
] ×
10−16
10−4
×
10
18
or Ey = 6.2 × 105Vm−1
So, magnitude of net electric field at point P is
EP = √Ex2+Ey2 = √2.32+6.22 × 105
or EP = 6.6 × 105Vm−1 directed at an angle
tan θ =
Ey
Ex
 =
6.2×105
−2.3×105
 = - 2.6956 or θ = - 69.6°
with the horizontal in –ve x-direction i.e. at 69.6° with BA.
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