JEE Advanced2025PhysicsElectrostaticsActual
A positive point charge of 10⁻⁸ C is kept at a distance of 20 cm from the center of a neutral conducting sphere of radius 10 cm . The sphere is then grounded and the charge on the sphere is measured. The grounding is then removed and subsequently the point charge is moved by a distance of 10 cm further away from the center of the sphere along the radial direction. Taking 1 4 ₀ =9 10^9 Nm ^2 / C ^2 (where ₀ is the per
Options
- ABefore the grounding, the electrostatic potential of the sphere is 450 V .
- BCharge flowing from the sphere to the ground because of grounding is 5 10⁻⁹ C .
- CAfter the grounding is removed, the charge on the sphere is -5 10⁻⁹ C .
- DThe final electrostatic potential of the sphere is 300 V .
Correct answer
A. Before the grounding, the electrostatic potential of the sphere is 450 V .
Step-by-step solution
Before grounding aligned & V _ sphere = ( V _ C )_ net = ( V _ C )_ q + ( V _ C )_ ind & V _ sphere = kq +0= 9 10^9 10⁻⁸ 0.2 = 90 0.2 = 900 2 =450 volt aligned After grounding aligned & kQ + kq _ S R = V _ sphere ^ =0 & q _ s =- R q =- 1 2 10⁻⁸=-5 10⁻⁹ & q _ s =-5 10⁻⁹ Coulomb aligned Charge flower from sphere to ground =5 10⁻⁹ Coulomb After grounding is removed aligned & ( V _ sphere )_ final = kq ^ + kq _ s R & = 9 10^9 10^2 10⁻⁸ 30 - 9 10^9 5 10⁻⁹ 10^2 10 & = 9 1000 30 -450=300 volt -450 volt =-150 volt aligned