Question

Two identical charged bodies have 12micro coulomb and
- 18micro coulombcharge resp .. These bodies experienced
a force of 48 N at certain seperation. The bodies
are touched and placed at same separation
again . find the new force between the bodies​

Answer 1

50 N

Explanation:

applying CISL find distance between them

now find the charge when kept in contact after that put all the required value in CISL

CISL(coulomb inverse square law)

Answer 2

The new force between the bodies​ is 2 N.

Explanation :

Given that,

Charge on first body = 12 μC

Charge on second body = -18 μC

Force F = 48 N

We need to calculate the force

Using formula of force

$F=\dfrac{1}{4\pi\epsilon_{0}}\times\dfrac{q_{1}\times q_{2}}{x^2}$

Put the value into the formula

$48=\dfrac{9\times10^{9}\times12\times10^{-6}\times18\times10^{-6}}{x^2}$...(I)

Two identical bodies having different magnitude of charge are touched, the redistribution of charge will be same on both bodies.

The new charge is

$q=\dfrac{q_{1}-q_{2}}{2}$

Put the value into the formula

$q=\dfrac{12-18}{2}$

$q=-3\ \mu C$

We need to calculate the new force between the bodies​

Using formula of force

$F=\dfrac{1}{4\pi\epsilon_{0}}\times\dfrac{q_{1}\times q_{2}}{x^2}$

Here, $q_{1}=q_{2}$

Put the value into the formula

$F=\dfrac{9\times10^{9}\times3\times10^{-6}\times3\times10^{-6}}{x^2}$...(II)

Divided equation (II) by equation(I)

$\dfrac{F}{48}=\dfrac{\dfrac{9\times10^{9}\times3\times10^{-6}\times3\times10^{-6}}{x^2}}{\dfrac{9\times10^{9}\times12\times10^{-6}\times18\times10^{-6}}{x^2}}$

$F=\dfrac{9\times48}{12\times18}$

$F=2\ N$

Hence, The new force between the bodies​ is 2 N.

Learn more :

Topic : electrostatic force

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