Question 170: Electrostatics & Capacitance

Question

A sphere of capacitance 100 pF is charged to a potential of 100 V . Another identical uncharged metal sphere is brought in contact with the charged sphere, then the change in the total energy stored on these spheres, when they touch is $\alpha 	imes 10^{-7} \mathrm{~J}$. The value of $\alpha$ is $\_\_\_\_$ .
(combined capacitance of spheres is 200 pF )

Accepted Answer

The work done in charging a capacitor is stored as electrostatic potential energy. If a charge dq is moved to a capacitor of capacitance C at potential $\mathrm{V}=\mathrm{q} / \mathrm{C}$, the work dW is:
$$ \mathrm{dW}=\mathrm{Vdq}=\frac{\mathrm{q}}{\mathrm{C}} \mathrm{dq} $$
Integrating from $\mathrm{q}=0$ to total charge $\mathrm{q}=\mathrm{Q}$ :
$$ \mathrm{U}=\int_0^{\mathrm{Q}} \frac{\mathrm{q}}{\mathrm{C}} \mathrm{dq}=\frac{1}{\mathrm{c}} \int_0^{\mathrm{Q}} \mathrm{qdq}=\frac{\mathrm{Q}^2}{2 \mathrm{C}} $$
Using $\mathrm{Q}=\mathrm{CV}$,
$$ \mathrm{U}=\frac{1}{2} \mathrm{CV}^2 $$
Initially, only the first sphere ( $\mathrm{C}_1=100 \mathrm{pF}$ ) is charged to $\mathrm{V}_1=100 \mathrm{~V}$.
$$ U_i=\frac{1}{2} C_1 V_1^2 $$
$$\Rightarrow $$ $U_i=\frac{1}{2} 	imes\left(100 	imes 10^{-12}ight) 	imes(100)^2$
$$\Rightarrow $$ $$ \mathrm{U}_{\mathrm{i}}=\frac{1}{2} 	imes 10^{-10} 	imes 10^4=0.5 	imes 10^{-6} \mathrm{~J}=5 	imes 10^{-7} \mathrm{~J} $$
Finally, when the spheres touch, they share charge until they reach a common potential $\mathrm{V}_{\mathrm{c}}$.
Since the spheres are identical ( $\mathrm{C}_1=\mathrm{C}_2=100 \mathrm{pF}$ ), the charge Q is shared equally, and
$$ \mathrm{V}_{\mathrm{c}}=\frac{	ext { Total Charge }}{	ext { Total Capacitance }}=\frac{\mathrm{C}_1 \mathrm{~V}_1+0}{\mathrm{C}_1+\mathrm{C}_2}=\frac{\mathrm{V}_1}{2}=50 \mathrm{~V} . $$
So, the final energy is,
$$ U_f=\frac{1}{2}\left(C_1+C_2ight) V_c^2 $$
$$\Rightarrow $$ $U_f=\frac{1}{2} 	imes\left(200 	imes 10^{-12}ight) 	imes(50)^2$
$$\Rightarrow $$ $$ U_f=100 	imes 10^{-12} 	imes 2500=2.5 	imes 10^{-7} J $$
So, the change in energy is,
$$ \Delta U=U_i-U_f $$
$$\Rightarrow $$ $\Delta \mathrm{U}=5 	imes 10^{-7} \mathrm{~J}-2.5 	imes 10^{-7} \mathrm{~J}=2.5 	imes 10^{-7} \mathrm{~J}=\alpha 	imes 10^{-7} \mathrm{~J}$
$\Rightarrow \alpha=2.5=\frac{5}{2}$

Answer

5

Answer

$\frac{5}{2}$

Answer

$\frac{7}{2}$

Answer

$\frac{9}{2}$

Electrostatics & Capacitance

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