A charge of $+ 2.0 \times 10^{- 8} C$ is placed on the positive plate and a charge of $- 1.0 \times 10^{- 8} C$ on the negative plate of a parallel-plate capacitor of capacitance $1.2 \times 10^{- 3} μ F .$

Calculate the potential difference developed between the plates.

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Solution

$q_{1} = 2.0 \times 10^{- 8} C,$

$q_{2} = - 1.0 \times 10^{- 8} C$

Net charge $= \frac{(q_{1} - q_{2})}{2}$

$= 1.5 \times 10^{- 8} C$

$q = V C$

$V = \frac{1.5 \times 10^{- 8}}{1.2 \times 10^{- 9}} = 12.5 V$

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A charge of +2.0×10−8 C is placed on the positive plate and a charge of −1.0×10−8 C on the negative plate of a parallel-plate capacitor of capacitance 1.2×10−3 μF. Calculate the potential difference developed between the plates.

q1=2.0×10−8 C, q2=−1.0×10−8 C Net charge =(q1−q2)2 =1.5×10−8 C q=VC V=1.5×10−81.2×10−9=12.5 V

A charge of $+ 2.0 \times 10^{- 8} C$ is placed on the positive plate and a charge of $- 1.0 \times 10^{- 8} C$ on the negative plate of a parallel-plate capacitor of capacitance $1.2 \times 10^{- 3} μ F .$

Calculate the potential difference developed between the plates.

$q_{1} = 2.0 \times 10^{- 8} C,$

$q_{2} = - 1.0 \times 10^{- 8} C$

Net charge $= \frac{(q_{1} - q_{2})}{2}$

$= 1.5 \times 10^{- 8} C$

$q = V C$

$V = \frac{1.5 \times 10^{- 8}}{1.2 \times 10^{- 9}} = 12.5 V$