Question 825: Electrostats & Capacitance

Question

A metallic sphere ' A ' isolated from ground is charged to $+50 \mu \mathrm{C}$. This sphere is brought in contact with other isolated metallic sphere ' $B$ ' of half the radius of sphere ' $A$ '. Then the charge on the two isolated spheres A \& B are in the ratio

Accepted Answer

When sphere A is brought into contact with sphere B, charge will transfer between them until their electrical potentials are equal. The relationship between charge and potential for a sphere is given by:
$ V = \frac{1}{4 \pi \varepsilon_0} \frac{q}{r} $
For both spheres A and B to have the same potential:
$ V_A = V_B $
Thus,
$ \frac{1}{4 \pi \varepsilon_0} \frac{q_A}{r_A} = \frac{1}{4 \pi \varepsilon_0} \frac{q_B}{r_B} $
This simplifies to:
$ \frac{q_A}{q_B} = \frac{r_A}{r_B} $
Given that sphere B has half the radius of sphere A:
$ r_B = \frac{r_A}{2} $
Substitute this into the equation:
$ \frac{q_A}{q_B} = \frac{r_A}{\frac{r_A}{2}} = \frac{2}{1} $
Hence, the charges on the spheres A and B are in the ratio $2:1$.

Answer

$1: 2$

Answer

$2: 1$

Answer

$4: 1$

Answer

$1: 1$

Electrostats & Capacitance

A metallic sphere ' A ' isolated from ground is charged to +50 \mu \mathrm{C}. This sphere is brought in contact with other isolated metallic sphere ' B ' of half the radius of sphere ' A '. Then the charge on the two isolated spheres A \& B are in the ratio

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