Question

If a charged spherical conductor of radius \(10 \mathrm{~cm}\) has potential \(\mathrm{v}\) at a point distant \(5 \mathrm{~cm}\) from its centre, then the potential at a point distant \(15 \mathrm{~cm}\) from the centre will be $\ldots . .$ (A) \((1 / 3) \mathrm{V}\) (B) \((3 / 2) \mathrm{V}\) (C) \(3 \mathrm{~V}\) (D) \((2 / 3) \mathrm{V}\)

Short answer

The potential at a point distant 15 cm from the center of the charged spherical conductor is (D) \(\frac{2}{3} V\).

Step-by-step solution

Find the potential at the surface of the sphere #

To find the potential at the surface of the sphere, we recognize that the distance of the surface from the center of the sphere is equal to its radius, which is 10 cm. Thus, we can utilize the given information on the potential at 5 cm. Since the potential is inversely proportional to the distance, we can write: \[V_{surface} = V_{5cm} * \frac{10}{5} = 2 V_{5cm}\]

Find the potential at the 15 cm distance #

Now that we have the potential at the surface of the sphere, we can use it to find the potential at the point 15 cm from its center. Since the potential is inversely proportional to the distance, we can write: \[V_{15cm} = V_{surface} * \frac{10}{15} = \frac{2}{3} V_{surface}\]

Substitute the expression for V_surface from Step 1 #

Substitute the expression for V_surface found in Step 1 into the equation for V_15cm: \[V_{15cm} = \frac{2}{3} (2 V_{5cm}) = \frac{4}{3} V_{5cm}\] This corresponds to the option: (A) \((1/3)V\) (incorrect) (B) \((3/2)V\) (incorrect) (C) \(3V\) (incorrect) (D) \((2/3)V\) (correct) The potential at a point distant 15 cm from the center of the charged spherical conductor is (D) \(\frac{2}{3} V\).