Question

How much charge should be placed on a capacitance of 25 pF to raise its potential
to 10^5V
(a)luc (b)1.5uC (c)2°C (d) 2.5uC
​

Answer 1

Answer:

(d) 2.5 uC

Explanation:

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Answer 2

Given:

Capacitor of 25 pF raises it potential to 10⁵ Volts.

To find:

Charge incorporated into capacitor?

Calculation:

The expression relating charge and potential difference in capacitor is :

$\therefore \: Q = C \times V$

$\implies \: Q = (25 \times {10}^{ - 12} ) \times ( {10}^{5} )$

$\implies \: Q = 25 \times {10}^{ - 7} \: Coulomb$

$\implies \: Q = 2.5 \times {10}^{ - 6} \: Coulomb$

$\implies \: Q = 2.5 \: \mu C$

So, charge needed is 2.5 micro-C.