Question

Surface charge density of a sphere of a radius 10 cm is 8.85x10-8C/m2. Potential at the centre of the
sphere is
1000 V
2) 885 V
3) 10-3 v
4) 442.5 V
​

Answer 1

Answer ⇒ Option (a). 1000 V.

Explanation ⇒ Given conditions,

Radius(r) = 10 cm.

= 0.1 m.

Surface charge density(σ) = 8.85 × 10⁻⁸ C/m².

Using the formula,

σ = charge/Surface area.

∴ 8.85 × 10⁻⁸ = Q/4π(0.1)²

∴ Q = 8.85 × 10⁻⁸ × 4π × R × R

∴ Q/4πR = 8.85 × 10⁻⁸ × R    ----eq(i)

Using the formula,

Potential = kQ/R, where R is the radius of the sphere.

∴ V = Q/4πε₀R

∴ V = Q/πR × 1/ε₀

∴ V =  8.85 × 10⁻⁸ × R ÷ 8.85 × 10⁻¹²

∴  V = 10⁴R

∴ V = 10000R

∴ V = 10000 × 0.1

∴ V = 1000 V.

Hence, option (a). is correct.

Hope it helps.

Answer 2

Answer ⇒ Option (a). 1000 V.

Explanation ⇒ Given conditions,

Radius(r) = 10 cm.

= 0.1 m.

Surface charge density(σ) = 8.85 × 10⁻⁸ C/m².

Using the formula,

σ = charge/Surface area.

∴ 8.85 × 10⁻⁸ = Q/4π(0.1)²

∴ Q = 8.85 × 10⁻⁸ × 4π × R × R

∴ Q/4πR = 8.85 × 10⁻⁸ × R    ----eq(i)

Using the formula,

Potential = kQ/R, where R is the radius of the sphere.

∴ V = Q/4πε₀R

∴ V = Q/πR × 1/ε₀

∴ V =  8.85 × 10⁻⁸ × R ÷ 8.85 × 10⁻¹²

∴  V = 10⁴R

∴ V = 10000R

∴ V = 10000 × 0.1

∴ V = 1000 V.