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Electric Field Due to a Sphere and Thin Shell

Two charges ...

Question

Two charges $q_{1} = 12 \times 10^{- 9} C$ and $q_{2} = - 12 \times 10^{- 9} C$ are placed $10 c m$ apart. The potential at a point $6 c m$ from $q_{1}$ on the line joining the two charges is :

3500 V o l t

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900 V o l t

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1800 V o l t

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- 900 V o l t

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Solution

The correct option is D $- 900 V o l t$
Total potential at A $= \frac{k q_{1}}{r_{1}} + \frac{k q_{2}}{r_{2}}$

$= \frac{k (12 \times 10^{- 9})}{6 \times 10^{- 2}} + \frac{k (- 12 \times 10^{- 9})}{4 \times 10^{- 2}}$

$= k (2 \times 10^{- 7}) + k (- 3 \times 10^{- 7})$
$= - 1 \times 10^{- 7} \times 9 \times 10^{9}$
$= - 900 V$

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Two charges q1=12×10−9C and q2=−12×10−9C are placed 10cm apart. The potential at a point 6cm from q1 on the line joining the two charges is :

The correct option is D −900VoltTotal potential at A =kq1r1+kq2r2=k(12×10−9)6×10−2+k(−12×10−9)4×10−2=k(2×10−7)+k(−3×10−7)=−1×10−7×9×109=−900V

Two charges $q_{1} = 12 \times 10^{- 9} C$ and $q_{2} = - 12 \times 10^{- 9} C$ are placed $10 c m$ apart. The potential at a point $6 c m$ from $q_{1}$ on the line joining the two charges is :