Question

Q1.two sphere of radius R and2R having same surface charge density. Two charges brought in contact then separated find the new surface charge density on each.​

Answer 1

Explanation:

Answer:

Surface Charge Density is represented as Charge per unit Area.

$\boxed{ \sigma = \dfrac{q}{A^2} } </p><p>$

According to the question, there are two spheres having radius 'R' and '2R' with same surface charge density.

⇒ S.D of Sphere 1 = S.D of Sphere 2

$\begin{gathered}\implies \dfrac{q_1}{\pi (R)^2} = \dfrac{q_2}{\pi (2R)^2}\\\\\\\implies \dfrac{q_1}{\pi (R)^2} = \dfrac{q_2}{4\pi(R)^2}\\\\\\\text{Cancelling the common terms we get:}\\\\\\\implies \dfrac{q_1}{1} = \dfrac{q_2}{4}\\\\\\\implies \boxed{ \bf{ 4q_1 = q_2}} \hspace{10} ...(1)\end{gathered}$

Now when both the spheres are brought close to each other, charges get transferred with each other. This happens when Potential Equilibrium is obtained. That is, when the electric potential of both the spheres is equal.

Electric Potential of a Sphere:

$\implies V_{sphere} = \dfrac{Kq}{r}$

Now let us assume that charges present in sphere 1 is Q₁ and charges present in sphere 2 is Q₂.

By the law of conservation of charge, charges cannot be created or destroyed. Hence we can say that:

⇒ q₁ + q₂ = Q₁ + Q₂

⇒ q₁ + 4q₁ = Q₁ + Q₂

⇒ 5q₁ = Q₁ + Q₂ ...(2)

Now calculating the potential of each sphere and equating it we get:

$</p><p>\begin{gathered}\implies \dfrac{KQ_1}{R} = \dfrac{KQ_2}{2R}\\\\\\\text{Cancelling the common terms we get:}\\\\\implies Q_! = \dfrac{Q_2}{2}\\\\\implies \boxed{ \bf{2Q_1 = Q_2}} \hspace{10} ...(3)\end{gathered}</p><p>$

Using (3) in (2) we get:

⇒ 5q₁ = Q₁ + 2Q₁

⇒ 5q₁ = 3Q₁

⇒ Q₁ = 5q₁/3

Therefore Q₂ is given as:

⇒ Q₂ = (2×5)q₁/3

⇒ Q₂ = 10q₁/3

Hence the new surface charge density of Sphere 1 is:

$\begin{gathered}\implies \dfrac{Q_1}{\pi R^2}\\\\\\\implies \dfrac{5}{3} \times \dfrac{q_1}{\pi R^2}\\\\\\\implies \boxed{\bf\sigma_{1}' = \dfrac{5}{3} \sigma}}\end{gathered}$

The new surface charge density of Sphere 2 is:

$\begin{gathered}\implies \dfrac{Q_2}{4 \pi R^2}\\\\\\\implies \dfrac{10}{3} \times \dfrac{q_2}{4 \pi R^2}\\\\\\\implies \boxed{ \bf{\sigma_{2}' = \dfrac{10}{3} \sigma}}\end{gathered} </p><p></p><p> </p><p> </p><p> </p><p> </p><p>′</p><p> </p><p> = </p><p>3</p><p>10</p><p> </p><p> σ</p><p> </p><p> </p><p>$

Answer 2

Answer:

Answer:

Surface Charge Density is represented as Charge per unit Area.

\boxed{ \sigma = \dfrac{q}{A^2} } < /p > < p >

σ=

A

2

q

</p><p>

According to the question, there are two spheres having radius 'R' and '2R' with same surface charge density.

⇒ S.D of Sphere 1 = S.D of Sphere 2

\begin{gathered}\begin{gathered}\implies \dfrac{q_1}{\pi (R)^2} = \dfrac{q_2}{\pi (2R)^2}\\\\\\\implies \dfrac{q_1}{\pi (R)^2} = \dfrac{q_2}{4\pi(R)^2}\\\\\\\text{Cancelling the common terms we get:}\\\\\\\implies \dfrac{q_1}{1} = \dfrac{q_2}{4}\\\\\\\implies \boxed{ \bf{ 4q_1 = q_2}} \hspace{10} ...(1)\end{gathered}\end{gathered}

Now when both the spheres are brought close to each other, charges get transferred with each other. This happens when Potential Equilibrium is obtained. That is, when the electric potential of both the spheres is equal.

Electric Potential of a Sphere:

\implies V_{sphere} = \dfrac{Kq}{r}⟹V

sphere

=

r

Kq

Now let us assume that charges present in sphere 1 is Q₁ and charges present in sphere 2 is Q₂.

By the law of conservation of charge, charges cannot be created or destroyed. Hence we can say that:

⇒ q₁ + q₂ = Q₁ + Q₂

⇒ q₁ + 4q₁ = Q₁ + Q₂

⇒ 5q₁ = Q₁ + Q₂ ...(2)

Now calculating the potential of each sphere and equating it we get:

\begin{gathered} < /p > < p > \begin{gathered}\implies \dfrac{KQ_1}{R} = \dfrac{KQ_2}{2R}\\\\\\\text{Cancelling the common terms we get:}\\\\\implies Q_! = \dfrac{Q_2}{2}\\\\\implies \boxed{ \bf{2Q_1 = Q_2}} \hspace{10} ...(3)\end{gathered} < /p > < p > \end{gathered}

Using (3) in (2) we get:

⇒ 5q₁ = Q₁ + 2Q₁

⇒ 5q₁ = 3Q₁

⇒ Q₁ = 5q₁/3

Therefore Q₂ is given as:

⇒ Q₂ = (2×5)q₁/3

⇒ Q₂ = 10q₁/3

Hence the new surface charge density of Sphere 1 is:

\begin{gathered}\begin{gathered}\implies \dfrac{Q_1}{\pi R^2}\\\\\\\implies \dfrac{5}{3} \times \dfrac{q_1}{\pi R^2}\\\\\\\implies \boxed{\bf\sigma_{1}' = \dfrac{5}{3} \sigma}}\end{gathered}\end{gathered}

The new surface charge density of Sphere 2 is:

\begin{gathered}\begin{gathered}\implies \dfrac{Q_2}{4 \pi R^2}\\\\\\\implies \dfrac{10}{3} \times \dfrac{q_2}{4 \pi R^2}\\\\\\\implies \boxed{ \bf{\sigma_{2}' = \dfrac{10}{3} \sigma}}\end{gathered} < /p > < p > < /p > < p > < /p > < p > < /p > < p > < /p > < p > < /p > < p > ′ < /p > < p > < /p > < p > = < /p > < p > 3 < /p > < p > 10 < /p > < p > < /p > < p > σ < /p > < p > < /p > < p > < /p > < p > \end{gathered}

⟹

4πR

2

Q

2

⟹

3

10

×

4πR

2

q

2

⟹

σ

2

′

=

3

10

σ

</p><p>

Explanation:

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