Question

Question:
A spherical drop water carrying a charge of 2
10-6 C has a potential of 500 V at
its surface. What is the radius of the drop ?​

Answer 1

The radius of the spherical water drop is equal to 36 metres.

Given:

The net charge on the spherical water drop = 2 × 10⁻⁶ C

The potential at the surface of the spherical water drop = 500 V

To Find:

The radius of the spherical water drop.

Solution:

→ The potential at the surface of a charged sphere is given by the formula:

                  $\boldsymbol{Electric\:Potential\:at\:Surface(V)=K\frac{Q}{R} }$

• where 'Q' is the net charge carried by the charged sphere.
• where 'R' is the radius of the charged sphere.
• where 'K' is constant having a value 9 × 10⁹ V m C⁻¹.

→ In the given question:

The net charge on the spherical water drop (Q) = 2 × 10⁻⁶ C

The potential at the surface of the spherical water drop (V) = 500 V

Let the radius of the spherical water drop be 'R'.

   $\boldsymbol{\because Electric\:Potential\:at\:Surface(V)=K\frac{Q}{R} }\\\\\boldsymbol{\therefore 500 = (9\times 10^{9} )\times \frac{2\times 10^{-6}}{R} }\\\\\boldsymbol{\therefore R=\frac{18000}{500}=36\:m }$

→ The value of 'R' comes out to be equal to 36 m.

Therefore the radius of the spherical water drop comes out to be equal to 36 metres.

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