Solveeit Logo

Question

Question: Let us assume that there are two identical conducting spheres M and N that have charges \({{q}_{m}}\...

Let us assume that there are two identical conducting spheres M and N that have charges qm{{q}_{m}} and qn{{q}_{n}} respectively. Let a third similar neutral sphere, P , be brought in contact with M and then separated. Now sphere P is brought in contact with N and then it is made separated. Then what will be the final charge on sphere P?
A.qm+2qn6 B.qm+qn4 C.qm+qn4 D.qm+2qn4 \begin{aligned} & A.\dfrac{{{q}_{m}}+2{{q}_{n}}}{6} \\\ & B.\dfrac{{{q}_{m}}+{{q}_{n}}}{4} \\\ & C.{{q}_{m}}+\dfrac{{{q}_{n}}}{4} \\\ & D.\dfrac{{{q}_{m}}+2{{q}_{n}}}{4} \\\ \end{aligned}

Explanation

Solution

First of all find the charge acquired by the sphere P when kept in contact with the sphere M. initially the charge on the sphere P will be zero. Then calculate the charge acquired by the sphere P when kept in contact with the sphere N. this time the initial charge will become equal to the charge acquired by the sphere P when kept in contact with sphere M.

Complete step by step solution:
The charge on the sphere M has been given as qm{{q}_{m}} and the charge on the sphere N has been given as qn{{q}_{n}}.
When the sphere P is brought in contact with the sphere M, then the charge on P will become,
qp=0+qm2{{q}_{p}}=\dfrac{0+{{q}_{m}}}{2}
Initially the charge of the sphere P will be zero.
After making the two spheres to be in contact, the charge on the sphere P will become,
qp=qm2{{q}_{p}}=\dfrac{{{q}_{m}}}{2}
Now the sphere P is brought in contact with N the amount charge will get varied in both the spheres. During this process initially the sphere P will be having a charge mentioned as,
qp=qm2{{q}_{p}}=\dfrac{{{q}_{m}}}{2}
And initially the charge in the sphere N will be mentioned as qn{{q}_{n}}.
Therefore after making these sphere in contact, the charge on the sphere P will become,
qp=qm2+qn2{{{q}'}_{p}}=\dfrac{\dfrac{{{q}_{m}}}{2}+{{q}_{n}}}{2}
Simplifying this will give,
qp=qm+2qn4{{{q}'}_{p}}=\dfrac{{{q}_{m}}+2{{q}_{n}}}{4}
Therefore the final charge on the sphere will become,
qp=qm+2qn4{{{q}'}_{p}}=\dfrac{{{q}_{m}}+2{{q}_{n}}}{4}
Hence the answer for the question has been calculated.

This answer has been given as option D.

Note:
Electric charge is defined as the basic physical characteristic of matter that will cause it to experience a force if it is placed in an electromagnetic field. There are two kinds of electric charges. One is classified as positive and the other is as negative.