An isolated sphere has a capacitance of 50pF. What would be the radius of the sphere?

An isolated sphere has a capacitance of 50pF. What would be the radius of the sphere?

(a) 90cm
(b) 45cm
(c) 11.50cm
(d) 5.75cm

45cm

As potential of spherical charge (Q) having radius r is given by V = (1 / 4πε₀) (Q / r)

Also, capacitance, C = Q / V = Q / { (1 / 4πε₀) (Q / r)} C = 4πε₀r

If C = 50 × 10⁻¹² F, then

Radius of the sphere r = 50 × 10⁻¹² / 4πε₀

= 50 × 10⁻¹² × 9 × 10⁹

= 50 × 9 × 10⁻³

= 450 × 10⁻³ m

= 0.450 m = 45 cm

UK2015

admin:

It's me says:

April 26, 2019 at 6:02 am

As Capacitance of Isolated sphere =4π€R
Given,
C=50picofarad
Then,
50×10(power -12)= 1/9×10(power9)×R
R=0.45m or 45cm....