A thin spherical shell with radius R_1 = 3.00 cm is concentric with a larger thi

Question

A thin spherical shell with radius R_1 = 3.00 cm is concentric with a larger thin spherical shell with radius R_2 = 5.00 cm. Both shells are made of insulating material. The smaller shell has charge q_1 = +6.00 nC distributed uniformly over its surface, and the larger shell has charge q_2 = - 9.00 nC distributed uniformly over its surface. Take the electric potential to be zero at an infinite distance from both shells, (a) What is the electric potential due to the two shells at the following distance from their common center: (i) r = 0; (ii) r = 4.00 cm; (iii) r = 6.00 cm? (b) What is the magnitude of the potential difference between the surfaces of the two shells? Which shell is at higher potential: the inner shell or the outer shell?

Explanation / Answer

here inner charge Qin = +6 nC

Outer charge Qout = -9nC

inner Radius R1 = 3 cm = 0.03 m

outer Radisu R2 = 5 cm = 0.05 m

Electric potential Due to apoint charge Q is V = KQ/r

where K is 9 *10^9 and r is the distance

so

part A :

At r =0 ,

Due to outer Shell, Electric Potential , Vout = KQout/R2

Vout = (9*10^9 * -9*10^-9)/(0.05)

Vout = - 1620 Volts

Due to inner Shell, Electric Potential , Vin = KQin/R1

Vin = (9*10^9 * -9*10^-9)/(0.03)

Vin = + 2700 Volts

Vnet at r = 0, is Vnet = Vin + Vout

Vnet at r =0 = 2700-1620

Vnet = 1080 Volts

------------------------------------

ii)

at r = 4cm

PD Vdiff = Vin - Vout

PD = K/r*(Qin - Qout)

Vnet = (9*10^9/0.04) *(6*10^-9-9*10^-9)

Vnet = -675 Volts

--------------------------------

iii)

at r= 6cm

Potential Diff = Vdiff = Vin - Vout

Vdiff = K/r*(Qin - Qout)

Vnet = (9*10^9/0.06) *(6*10^-9-9*10^-9)

Vnet = -450 Volts

----------------------

Since inner shell is +ve

it is at Higher Potential

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