Charges of 5 times 10-6- and -3 times 10-6- C are kept 16 cm apart- Find the position of points on the line joining the charges where the electric potential is zero-

Question

Toppr Admin · Accepted Answer

Potential at P iss zerok-5-xD7-10-x2212-6-x-k-3-xD7-10-x2212-6-0-16-x2212-x5-0-16-x2212-x-3x0-8-8xx-0-1mx-10cm

Charges of $5 \times 10^{- 6}$ and $- 3 \times 10^{- 6} C$ are kept 16 cm apart. Find the position of points on the line joining the charges where the electric potential is zero.

Potential at P iss zero
$\frac{k (5 \times 10^{- 6})}{x} = \frac{k (3 \times 10^{- 6})}{0.16 - x}$

$5 (0.16 - x) = 3 x$
$0.8 = 8 x$
$x = 0.1 m$
$x = 10 c m$

Charges of 5×10−6 and −3×10−6C are kept 16 cm apart. Find the position of points on the line joining the charges where the electric potential is zero.

Potential at P iss zerok(5×10−6)x=k(3×10−6)0.16−x5(0.16−x)=3x0.8=8xx=0.1mx=10cm

Charges of $5 \times 10^{- 6}$ and $- 3 \times 10^{- 6} C$ are kept 16 cm apart. Find the position of points on the line joining the charges where the electric potential is zero.

Potential at P iss zero
$\frac{k (5 \times 10^{- 6})}{x} = \frac{k (3 \times 10^{- 6})}{0.16 - x}$

$5 (0.16 - x) = 3 x$
$0.8 = 8 x$
$x = 0.1 m$
$x = 10 c m$