Question

derive the electrical energy stored in the system due to the particle system.​

Answer 1

Answer:

Potential energy of dipole in uniform electric field.

τ

=

p

×

E

work done in rotating dipole by small angle

dw=τdθ

=−pεsinθdθ

The change in potential energy

dU=−dW=pEsinθdθ

If θ→90

o

to θ

o

U(θ)−U(90

o

)=∫

90

o

0

o

Pεsinθdθ

=Pε[−cosθ]

90

0

=−

P

.

E

So

U(θ)=−

P

.

E

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Answer 2

To derive:

Electrostatic Potential Energy stored in a binary charge system.

Derivation:

Let the charges be q and Q separated by a distance d;

Potential of charge q at the location of Q:

$\displaystyle \: \int \: dV = - \int E \times dr$

$= > \displaystyle \: \int \: dV = - \int \dfrac{kq}{ {r}^{2} } \times dr$

Putting limits:

$= > \displaystyle \: \int_{0}^{V} \: dV = - \int_{ \infty}^{d} \dfrac{kq}{ {r}^{2} } \times dr$

$= > \displaystyle \: V = - \int_{ \infty}^{d} \dfrac{kq}{ {r}^{2} } \times dr$

$= > \displaystyle \: V = - kq \int_{ \infty}^{d} \dfrac{dr}{ {r}^{2} }$

$= > \displaystyle \: V = kq \bigg \{ \frac{1}{d} - \frac{1}{ \infty} \bigg \}$

$= > \displaystyle \: V = kq \bigg \{ \frac{1}{d} - 0 \bigg \}$

$= > \displaystyle \: V = \frac{kq}{d}$

So, Potential Energy stored in the system

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

$= > \: U = \dfrac{kq}{d} \times Q$

$= > \: U = \dfrac{kqQ}{d}$

So, final answer is:

$\boxed{ \bf{ \large{\: U = \dfrac{kqQ}{d} }}}$