Question 603: Electrostatics & Capacitance

Question

A $$12 \mathrm{~pF}$$ capacitor is connected to a $$50 \mathrm{~V}$$ battery, the electrostatic energy stored in the capacitor in $$\mathrm{nJ}$$ is

Accepted Answer

The electrostatic energy stored in a capacitor can be calculated using the formula:
$$ U = \frac{1}{2} C V^2 $$
Where:
$$ U $$ is the stored energy in joules (J)
$$ C $$ is the capacitance in farads (F)
$$ V $$ is the voltage in volts (V)
Given:
$$ C = 12 \mathrm{~pF} $$ (which is $$ 12 	imes 10^{-12} $$ F)
$$ V = 50 \mathrm{~V} $$
Substitute the values into the formula:
$$ U = \frac{1}{2} 	imes 12 	imes 10^{-12} 	imes (50)^2 $$
Calculate the values inside the parentheses first:
$$ 50^2 = 2500 $$
Now substitute this back into the equation:
$$ U = \frac{1}{2} 	imes 12 	imes 10^{-12} 	imes 2500 $$
Perform the multiplication:
$$ U = \frac{1}{2} 	imes 30000 	imes 10^{-12} $$
Which simplifies to:
$$ U = 15000 	imes 10^{-12} $$
Convert this to nanojoules (nJ) by recognizing that $$ 1 \mathrm{~nJ} = 10^{-9} \mathrm{~J} $$:
$$ U = 15 \mathrm{~nJ} $$
Thus, the electrostatic energy stored in the capacitor is:
Option A: 15 nJ

Answer

15

Answer

7.5

Answer

0.3

Answer

150

Electrostatics & Capacitance

A $12 \mathrm{~pF} capacitor is connected to a 50 \mathrm{~V} battery, the electrostatic energy stored in the capacitor in \mathrm{nJ}$ is

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