Question

establish on expression for electric energy of a capacitor is capacity 'c' changed to potential 'v'.​

Answer 1

ENERGY STORED IN CAPACITOR:

• Let's assume that battery does a small work dW in delivering a small charge dq at a constant voltage of V.

$\rm dW = V \times dq$

$\rm \implies dW = \dfrac{q}{C} \times dq$

$\rm \implies dW = \dfrac{1}{C} \times q \: dq$

Now , integrating on both sides:

$\rm \implies \displaystyle \int_{0}^{W}dW = \dfrac{1}{C} \times \int_{0}^{q} q \: dq$

$\rm \implies W = \dfrac{1}{C} \times \dfrac{ {q}^{2} }{2}$

$\rm \implies W = \dfrac{1}{C} \times \dfrac{ {(CV)}^{2} }{2}$

$\rm \implies W = \dfrac{ {C(V)}^{2} }{2}$

So, energy stored in capacitor is CV²/2.

Answer 2

Explanation:

ENERGY STORED IN CAPACITOR:

Let's assume that battery does a small work dW in delivering a small charge dq at a constant voltage of V.

\rm dW = V \times dqdW=V×dq

\rm \implies dW = \dfrac{q}{C} \times dq⟹dW=

C

q

×dq

\rm \implies dW = \dfrac{1}{C} \times q \: dq⟹dW=

C

1

×qdq

Now , integrating on both sides:

\rm \implies \displaystyle \int_{0}^{W}dW = \dfrac{1}{C} \times \int_{0}^{q} q \: dq⟹∫

0

W

dW=

C

1

×∫

0

q

qdq

\rm \implies W = \dfrac{1}{C} \times \dfrac{ {q}^{2} }{2} ⟹W=

C

1

×

2

q

2

\rm \implies W = \dfrac{1}{C} \times \dfrac{ {(CV)}^{2} }{2} ⟹W=

C

1

×

2

(CV)

2

\rm \implies W = \dfrac{ {C(V)}^{2} }{2} ⟹W=

2

C(V)

2

So, energy stored in capacitor is CV²/2.✳️✳️