Question

Two charges 3 x 10-8 C and -2 x 10-8 Care located 15 cm apart. At what point on the line
joining the two charges is the electric potential zero? Take the potential at infinity to be zero​

Answer 1

Concept Introduction: Charges follow Coulomb's Law.

Given:

We have been Given: Magnitude of Charge one,

$3 \times {10}^{ - 8} c$

Magnitude of Charge two is,

$- 2 \times {10}^{ - 8} c$

Distance d between the charges,

$15 \: cm = 0.15m$

To Find:

We have to Find: At what point on the line

joining the two charges is the electric potential zero?

Solution:

According to the problem, Let the point at which the potential is zero be P , distance from the point of zero potential to a charge be r.

Therefore using the formula,

$v = \frac{1}{4\pi \times e} \times \frac{q1}{r} + \frac{1}{4\pi \times e} \times \frac{q2}{(d - r)}$

Now,

$v = 0$

therefore,

$\frac{q1}{r} = - \frac{q2}{(d - r)}$

Putting the values in the place,

$\frac{3 \times {10}^{ - 8} }{r} = - \frac{ - 2 \times {10}^{ - 8} }{(0.15 - r)} \\ \frac{3}{r} = \frac{2}{(0.15 - r)} \\ 3(0.15 - r) = 2r \\ 0.45 - 3r = 2r \\ 3r + 2r = 0.45 \\ 5r = 0.45 \\ r = 0.09m = 9 \: cm$

Final Answer:

Therefore, The potential is zero at

$9cm$

on the line joining the charges.

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