Question

Two charged metallic spheres of radii, R = 10 cms and R2 = 20 cms are touching each other. If the charge on each sphere is +100 nC, what is the electric potential energy between the two charged spheres?

Answer 1

The electric potential energy between the two charged spheres is
`199.9*10^(-6)\ J`

Step-by-step explanation:

Given that,

Radius of first sphere
`R_(1)=10\ cm`

Radius of second sphere
`R_(2)=10\ cm`

Charge Q= 100 nC

We know charge flows through higher potential to lower potential.

Using formula of potential

`V=(k(Q-q))/(R_(1))`...(I)

`V=(k(Q+q))/(R_(2))`...(II)

From equation (I) and (II)

`(k(Q-q))/(R_(1))=(k(Q+q))/(R_(2))`

Put the value into the formula

`((100-q))/(10*10^(2))=((100+q))/(20*10^(-2))`

`(100-q)*20*10^(-2)=(100+q)*10*10^(-2)`

`q=(1000)/(30)`

`q=(100)/(3)`

`q=33.33\ nC`

So, the potential at R₁ and R₂

Using formula of potential

`V=(k(Q-q))/(R_(1))`

Put the value into the formula

`V=(9*10^(9)(100-33.33)*10^(-9))/(10*10^(-2))`

`V=6000.3\ Volt`

We need to calculate the electric potential energy between the two charged spheres

Using formula of the electric potential energy

`U=qV`

`U=33.33*10^(-9)*6.0003*10^(3)`

`U=199.9*10^(-6)\ J`

Hence, The electric potential energy between the two charged spheres is
`199.9*10^(-6)\ J`

Answer 2

`200* 10^(-6)j`

Step-by-step explanation:

We have given the radius of first sphere is 10 cm and radius of second sphere is 20 cm

So the potential of first sphere will be greater than the potential of the second sphere, so charge will flow from first sphere to second sphere

Let q charge is flow from first sphere to second sphere and then potential become same

So
`V=(K(100-q))/(r_1)=(K* 100)/(r_2)`

200-100=2q+q

`q=(100)/(3)=33.33nC`

So
`V=(K(100-q))/(r_1)=(9* 10^(9)* (100-33.33)* 10^(-9))/(10* 10^(-2))=6003V`

We know that potential energy U=qV
`=33.33* 10^(-9)* 6003=200* 10^(-6)j`