Question

A large isolated metal sphere of radius r carries a fixed charge. A small charge is placed at a distance s from its surface. It experiences a force which is
a) inversely proportional to (r + s)
b) inversely proportional to s
c) inversely proportional to s2
d) inversely proportional to (r + s)2​

Answer 1

Answer:

A large isolated metal sphere of radius r carries a fixed charge. A small charge is placed at a distance s from its surface. It experiences a force which is inversely proportional to s.

Explanation:

According to question , a large isolated metal sphere of radius r carries a fixed charge. A small charge is placed at a distance s from its surface.It experiences a force. we need to find the relation of the force with the distance s.

There are 4 options here.

The correct option is b) inversely proportional to s

So the answer is

A large isolated metal sphere of radius r carries a fixed charge. A small charge is placed at a distance s from its surface. It experiences a force which is inversely proportional to s.

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Answer 2

Answer:

The charge experiences a force that is inversely proportional to (r+s)².

Explanation:

• Coulomb's law states that the force of attraction or repulsion between two charged bodies is inversely proportional to the square of the distance between them and directly proportional to the product of their charges.
• It acts along the line that connects the two charges that are regarded as point charges.

        $F$ ∝ $\frac{q_{1}q_{2} }{r^{2} }$

        $F = k \frac{q_{1}q_{2} }{r^{2} }$

Where,

$k$ ⇒ 1/4πε₀ = 9 × 10⁹

Let r be the radius of the sphere and q₀ be the charge placed at the distance s to the metal sphere.

The total distance = r+s

Now from Coulomb's law,

$F = k \frac{q_{1}q_{o} }{(r+s)^{2} }$

Therefore,v

$F$ ∝ $\frac{1}{(r+s)^{2} }$

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