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Standard XII

Physics

Equipotential Surfaces

Two point cha...

Question

Two point charges, $q_{1} = 15 \times 10^{- 8}$ C and $q_{2} = - 3 \times 10^{8}$ C,are separated by a distance of 60 cm in air.Find the position of point situated between the two charges and at a distance from $q_{2}$ at which the electric potential would be zero

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Solution

Potential at P:
$V = \frac{K \times 15 \times 10^{- 8}}{60 - x} + \frac{- 3 \times k \times 10^{- 8}}{x} = 0$
$\frac{15 k x \times 10^{- 8} - 3 k (60 - x) \times 10^{- 8}}{60 x - x^{2}} = 0$
$15 k x - 180 k + 3 k x = 0$
$x = 10 c m$

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Two point charges,q1=15×10−8 C and q2=−3×108 C,are separated by a distance of 60 cm in air.Find the position of point situated between the two charges and at a distance from q2 at which the electric potential would be zero

Potential at P:V=K×15×10−860−x+−3×k×10−8x=015kx×10−8−3k(60−x)×10−860x−x2=015kx−180k+3kx=0x=10cm

Two point charges, $q_{1} = 15 \times 10^{- 8}$ C and $q_{2} = - 3 \times 10^{8}$ C,are separated by a distance of 60 cm in air.Find the position of point situated between the two charges and at a distance from $q_{2}$ at which the electric potential would be zero

Potential at P:
$V = \frac{K \times 15 \times 10^{- 8}}{60 - x} + \frac{- 3 \times k \times 10^{- 8}}{x} = 0$
$\frac{15 k x \times 10^{- 8} - 3 k (60 - x) \times 10^{- 8}}{60 x - x^{2}} = 0$
$15 k x - 180 k + 3 k x = 0$
$x = 10 c m$