Question

the radii of two charged metallic spheres are 5 cm and 10 cm . each has a charge of 75 microcoulomb. they are connected by a conducting wire . calculate:
1. common potential of the spheres after connecting
2. amount of charge transferred

Answer 1

answer : 1. 9 × 10^6 volts

2. 25$\mu C$

explanation : it is given that, the radii of two charged metallic spheres are 5 cm and 10 cm and each has a charge of 75$\mu C$

so, initial potential of first sphere, $v_1=\frac{kQ}{r_1}$

and initial potential of 2nd sphere, $v_2=\frac{kQ}{r_2}$

since, $r_1< r_2$ potential of first sphere is higher than potential of 2nd sphere and we know, charge flows through higher potential to lower potential so, charge flows through first sphere to 2nd sphere

Let q charge transferred from first sphere to 2nd sphere after some time, both spheres attain same potential. e.g., V.

so, V = $\frac{k(Q-q)}{r_1}=\frac{K(Q+q)}{r_2}$

or, $\frac{75\mu C-q}{5cm}=\frac{75\mu C+q}{10cm}$

or, q = 25 $\mu C$

hence, amount of charge transferred is 25$\mu C$

common potential, V = K(Q -q)/r1

= 9 × 10^9 × (75 - 25) × 10^-6/0.05

= 9 × 10^6 volts

hence, common potential of the spheres after connecting = 9 × 10^6 volts

Answer 2

Answer:

Explanation:

Radius of first sphere= 5cm

Radius of second sphere = 10cm

Charge = 75mcb

The initial potential of first sphere, v1 = kq/r1

The initial potential of second sphere =  v2 = kq/r2

Since,  r1 < r2 the potential of first sphere will be higher than the potential of second sphere. Since we know that the charge flows through higher potential to lower potential, thus the charge flows through first sphere to second sphere.

Let the charge q is transferred from first sphere to second sphere, thus after some time, both spheres attain same potential. e.g., W. Thus,

W = k (Q-q)/r1 = k (Q+q)/r2

W = 75-q/5 = 75+q/10

W = 25

Thus, the amount of charge transferred is 25

Common potential, V = K(Q -q)/r1

= 9 × 10^9 × (75 - 25) × 10^-6/0.05

= 9 × 10^6

Therefore, the common potential of the spheres after connecting = 9 × 10^6 volts.