Two charges 5 x 10-8 C and – 3 x 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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1 Answer
On the line there are two locations where potential becomes zero.
For P,
\(\frac{5 \times 10^{-8}}{x}\) = \(\frac{3 \times 10^{-8}}{16-x}\)
5( 16 - x) = 3x
80 - 5x = 3x
8x = 80
∴ x = 10cm
For Q,
\(\frac{5 \times 10^{-8}}{16+y}\) = \(\frac{3 \times 10^{-8}}{y}\)
5y = 48 + 3y
∴ 2y = 48
y = 24 cm
i.e., at (16 + 24) = 40 cm from A
For P,
\(\frac{5 \times 10^{-8}}{x}\) = \(\frac{3 \times 10^{-8}}{16-x}\)
5( 16 - x) = 3x
80 - 5x = 3x
8x = 80
∴ x = 10cm
For Q,
\(\frac{5 \times 10^{-8}}{16+y}\) = \(\frac{3 \times 10^{-8}}{y}\)
5y = 48 + 3y
∴ 2y = 48
y = 24 cm
i.e., at (16 + 24) = 40 cm from A