SATHEE: Electrostatics 5 Question 6

Electrostatics 5 Question 6

7. A parallel plate capacitor has $1 μ F$ capacitance. One of its two plates is given $+ 2 μ C$ charge and the other plate $+ 4 μ C$ charge. The potential difference developed across the capacitor is

(Main 2019, 8 April II)

(a) $1 V$

(b) $5 V$

(c) $2 V$

(d) $3 V$

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Answer:

Correct Answer: 7. (a)

Solution:

Net value of charge on plates of capacitor after steady state is reached is

$q_{net} = \frac{q_{2} - q_{1}}{2}$

where, $q_{2}$ and $q_{1}$ are the charges given to plates.

(Note that this formula is valid for any polarity of charge.)

Here, $q_{2} = 4 μ C, q_{1} = 2 μ C$

$∴$ Charge of capacitor is $q = Δ q_{net} = \frac{4 - 2}{2} = 1 μ C$

Potential difference between capacitor plates is

$V = \frac{Q}{C} = \frac{1 μ C}{1 μ F} = 1 V$

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Electrostatics 5 Question 6

7. A parallel plate capacitor has 1μF capacitance. One of its two plates is given +2μC charge and the other plate +4μC charge. The potential difference developed across the capacitor is

Answer:

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7. A parallel plate capacitor has $1 μ F$ capacitance. One of its two plates is given $+ 2 μ C$ charge and the other plate $+ 4 μ C$ charge. The potential difference developed across the capacitor is