Question

Two metal spheres, one of radius R and the other of radius $2R$, both have same surface charge density. They are brought in contact and separated. What will be new surface charge densities on them?

Answer 1

Formula used: $\sigma =qA,V=KqR$

Given,

Radius of first sphere =$R$

Radius of second sphere =$2R$

Surface charge density on both spheres =$\sigma$

Let the charge on sphere with radius R is $q_1$ and charge on sphere with radius 2R is $q_2$.

Before contact, charges of each sphere,

$q_1=\sigma 4\pi R^2$

$q_2=\sigma 4\pi (2R{)}^2=4q_1$

When the two sphere are brought in contact, their potentials become equal i.e., $V_1=V_2$. Let the charges on the spheres after getting in contact be $q_1$ and $q_2$ therefore

$\frac{q_1}{4\pi {ϵ}_{0}R}=\frac{q_2}{4\pi {ϵ}_0\left(2R\right)}$

$\therefore q_2=2q_1$ …(i)

As there is no loss of charge in the process.
Apply conservation of charge,

$q_1^{\prime }+q_2^{\prime }=q_1+q_2=q_1+4q_1=5q_1=5\sigma 4\pi R^2$

$q_1^{\prime }+2q_1^{\prime }=5\sigma 4\pi R^2$ (using (i))

$q_1^{\prime }=\frac{5}{3}4R^2$

$\therefore q_2^{\prime }=2q_1^{\prime }$

$q_2^{\prime }=\frac{10}{3}4R^2$

$1=\frac{q_1}{4\pi R^2}$ put the value of $q_1^{\prime }=\frac{5}{3}\sigma$

$2=\frac{q_2}{4\pi (2R{)}^2}$ put the value of$q_2^{\prime }=\frac{5}{6}\sigma$

Final Answer: $\frac{5}{3}\sigma ,\frac{5}{6}\sigma$