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Question
Two charges 5 × 10−8 C and −3 × 10−8 C are located 16 cm apart. At what point(s) on the line joining the two charges is the electric potential zero? Take the potential at infinity to be zero.
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Solution
Given: q1 = 5 × 10−8 C
q2 = −3 × 10−8 C
r = 16 cm
Let C be the point on the line joining the two charges, where the electric potential is zero, then
VC = 0
VCA + VCB = 0
VCA = −VCB
`1/(4piε_0) q_A/r_(CA) = -1/(4piε_0) q_B/r_(CB)`
`(5 xx 10^-8 C)/(x xx 10^-2 m) = (-(-3 xx 10^-8 C))/([(16 - x) xx 10^-2 m])`
`5/x = 3/(16 -x)`
5(16 − x) = 3x
80 − 5x = 3x
8x = 80 m
∴ `x = 80/8`
= 10 cm
So, the electric potential is zero at a distance of 10 cm from the charge 5 × 10−8 C on the line joining the two charges, between them.
If point C is not between the two charges, then
VCA + VCB = 0
VCA = −VCB
`1/(4piε_0) q_A/r_(CA) = -1/(4piε_0) q_B/r_(CB)`
`(5 xx 10^-8 C)/([(16 + x) xx 10^-2 m]) = (-(-3 xx 10^-8 C))/(x xx 10^-2 m)`
`5/(16 + x) = 3/x`
5x = 48 + 3x
2x = 48 m
= 24 cm
So, the electric potential is also zero at a distance of 24 cm from the charge −3 × 10−8 C, i.e., at a distance of (24 + 16) = 40 cm from the charge 5 × 10−8 C on the side of the charge −3 × 10−8 C.
RELATED QUESTIONS
An electric dipole is held in a uniform electric field.
(i) Show that the net force acting on it is zero.
(ii) The dipole is aligned parallel to the field. Find the work done in rotating it through the angle of 180°.
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Assertion (A): The potential (V) at any axial point, at 2 m distance (r) from the centre of the dipole of dipole moment vector `vec P` of magnitude, 4 × 10−6 cm, is ±9 × 103 V.
(Take `1/(4 pi epsilon_0)` = 9 × 109 SI units)
Reason (R): V = `+- (2 vec P)/(4 pi epsilon_0 r^2)`, where r is the distance of any axial point, situated at 2 m from the centre of the dipole.
In the light of the above statements, choose the correct answer from the options given below:
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The exact potential at a general point of a dipole with half-length l, before neglecting the \[l^2\cos^2\theta\] term, is:
In the general formula for the potential due to a dipole, setting θ = 0° and θ = 90° respectively recovers which two special cases?
For a dipole with \[p = 3 \times 10^{-10}\ \text{C}\cdot\text{m}\], \[l = 0.05\ \text{m}\], evaluated at an axial point \[r = 0.2\ \text{m}\] using \[k = 9 \times 10^9\ \text{N}\cdot\text{m}^2/\text{C}^2\] in the formula \[V = \frac{kp}{r^2 - l^2}\], the potential is:
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