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11 Electrostatics
11.98 Both gravitational field and electric field obey inverse square law. Using this
analogy show that the differential Gauss’ law for gravitation is given by ∇ ·
g = −ρ/G, where ρ is the mass density.
11.3 Solutions
11.3.1 Electric Field and Potential
11.1 (a)
(i) F =
q 1 q 2
4πε 0 r 2 =
(8 × 1.6 × 10 −19 )(−4 × 1.6 × 10 −19 )
4π × 8.85 × 10 −12 × (0.2) 2
= −1.8432 × 10
−25 N
(ii) For the position of zero electric field the forces due to the two charges
must be equal in magnitude but oppositely directed. Clearly the neutral point must be on the x-axis. On the left of Q 1 , the forces will
be oppositely directed but cannot be equal as |Q 1 | > |Q 2 |. Between
Q 1 and Q 2 , the forces are exerted in the same direction. On the right
of Q 2 conditions are favourable for a null point. Let the zero electric
field be situated at a distance x from Q 2 on the right.
8e
(x + 0.2) 2 −
4e
x 2 = 0
whence x = 0.4828 on the right of Q 2 .
(b) E =
e
4 πε 0 r 2 =
1.6 × 10 −19
4π × 8.85 × 10 −12 × (5.3 × 10 −11 ) 2
= 5.12 × 10 18 N/C
Force F = Ee = 5.12 × 10 18 × 1.6 × 10 −19 = 81.92 N
11.2 (a) As the electric field is downwards, the force on the positive charge will
be downwards and the force on the negative charge will be upwards.
(i) q = +8 μC
F q = q E = +8 × 10
−6
× 300 = 2.4 × 10
−3 N
F g = mg = 0.6 × 10
−3
× 9.8 = 5.88 × 10
−3 N
∴ Tension in the thread
T = F g + F q = 5.88 × 10
−3
+ 2.4 × 10
−3
= 8.28 × 10
−3 N