Question

Two metal spheres a and b of radius r and 2r centres are separated by a distance r are given a charge q and r at a potential v1 and v2 find the ratio of v1/v2

Answer 1

Answer:

Explanation:

V1 = kq/r + kq/6r = 7q/6r

V2= kq/2r+kq/6r = 4kq/6r

therefore, V1/V2 = 7/4

V(commom) =2kq/(r+2r) =V'

charge transferred is equal to  

q'= C1V1- C1V' = (r/k * kq/r) -(r/k * k2q/3r)  = q - 2q/3  = q/3

Answer 2

The ratio of V1/V2 is 2:1.

Explanation:

The potential of a sphere of radius r carrying a charge q is  

$V=\frac{kq}{r}$

Here, q is the charge and r is the radius.

For first metal sphere,

$V_1=\frac{kq}{r}$

For second,

$V_2=\frac{kq}{2r}$

Therefore,

$\frac{V_1}{V_2} =\frac{\frac{kq}{r} }{\frac{kq}{2r} }$

$\frac{V_1}{V_2} =\frac{2}{1}$

Thus, the ratio of V1/V2 is 2:1.

#Learn More: Potential

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