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Question: The sum of two-point charges is 7microC. They repel each other with a force of 1N when kept 30cm apa...

The sum of two-point charges is 7microC. They repel each other with a force of 1N when kept 30cm apart in free space. Calculate the value of each charge.
(A)2μC and 4μC (B)2μC and 5μC (C)9μC and 5μC (D)4μC and 7μC \begin{aligned} & \left( A \right)2\mu C\text{ and 4}\mu C \\\ & \left( B \right)2\mu C\text{ and 5}\mu C \\\ & \left( C \right)9\mu C\text{ and 5}\mu C \\\ & \left( D \right)4\mu C\text{ and 7}\mu C \\\ \end{aligned}

Explanation

Solution

The sum of these two charges is equal to 7microC which is a positive entity. And since, the nature of force is repelling in nature, the charges must be positive. We will take these two charges as two separate variables and then construct two different linear equations to find the value of these two variables.

Complete answer:
Let us first assign some terms that we are going to use later in our problem.
Let the two charges be given by ‘+p+p’ and ‘+q+q’. Then, according to the first statement in the problem, we have:
⇒p+q=7μC ⇒p+q=7×10−6C \begin{aligned} & \Rightarrow p+q=7\mu C \\\ & \Rightarrow p+q=7\times {{10}^{-6}}C \\\ \end{aligned}
⇒p=(7×10−6−q)C\Rightarrow p=\left( 7\times {{10}^{-6}}-q \right)C .......... (1)
Now, according to the second statement in our problem, the net force of repulsion between the charges is of 1N. Using the expression of electrostatic force to write the force of repulsion between these two charges, we get:
⇒F=kq1q2r2\Rightarrow F=k\dfrac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}
Where,
‘F’ is the electrostatic force of repulsion which is equal to 1N.
q1 and q2{{q}_{1}}\text{ and }{{q}_{2}} are the two charges.
‘r’ is the distance between the two charges which is equal to 30cm. And,
‘k’ is the Coulomb’s constant which is equal to 9×109Nm2C−29\times {{10}^{9}}N{{m}^{2}}{{C}^{-2}} .
Putting the values of these terms in the above force equation, we get:
⇒1=9×109p×q(30×10−2)2 ⇒1=9×109p×q900×10−4 ⇒1=1011(pq) ⇒pq=10−11 \begin{aligned} & \Rightarrow 1=9\times {{10}^{9}}\dfrac{p\times q}{{{\left( 30\times {{10}^{-2}} \right)}^{2}}} \\\ & \Rightarrow 1=9\times {{10}^{9}}\dfrac{p\times q}{900\times {{10}^{-4}}} \\\ & \Rightarrow 1={{10}^{11}}\left( pq \right) \\\ & \Rightarrow pq={{10}^{-11}} \\\ \end{aligned}
Using the value of ‘p’ from equation number (1), we get:
⇒q(7×10−6−q)=10−11 ⇒q2−7×10−6q+10−11=0 ⇒q2−5×10−6q−2×10−6q+10−11=0 ⇒q(q−5×10−6)−2×10−6(q−5×10−6)=0 ⇒(q−5×10−6)(q−2×10−6)=0 ∴q=5×10−6C or 2×10−6C \begin{aligned} & \Rightarrow q\left( 7\times {{10}^{-6}}-q \right)={{10}^{-11}} \\\ & \Rightarrow {{q}^{2}}-7\times {{10}^{-6}}q+{{10}^{-11}}=0 \\\ & \Rightarrow {{q}^{2}}-5\times {{10}^{-6}}q-2\times {{10}^{-6}}q+{{10}^{-11}}=0 \\\ & \Rightarrow q\left( q-5\times {{10}^{-6}} \right)-2\times {{10}^{-6}}\left( q-5\times {{10}^{-6}} \right)=0 \\\ & \Rightarrow \left( q-5\times {{10}^{-6}} \right)\left( q-2\times {{10}^{-6}} \right)=0 \\\ & \therefore q=5\times {{10}^{-6}}C\text{ or }2\times {{10}^{-6}}C \\\ \end{aligned}
Thus, the corresponding value of ‘p’ comes out to be:
⇒p=(7×10−6−q)C ⇒p=(7×10−6−5×10−6)C or, p=(7×10−6−2×10−6)C ∴p=2×10−6C or 5×10−6C \begin{aligned} & \Rightarrow p=\left( 7\times {{10}^{-6}}-q \right)C \\\ & \Rightarrow p=\left( 7\times {{10}^{-6}}-5\times {{10}^{-6}} \right)C\text{ or, }p=\left( 7\times {{10}^{-6}}-2\times {{10}^{-6}} \right)C \\\ & \therefore p=2\times {{10}^{-6}}C\text{ or 5}\times {{10}^{-6}}C \\\ \end{aligned}
Assuming pp<qq, we get our solution in ‘p’ and ‘q’ as 2μC and 5μC\text{2}\mu \text{C and 5}\mu \text{C}.
Hence, the value of each charge comes out to be 2μC and 5μC\text{2}\mu \text{C and 5}\mu \text{C}.

Hence, option (B) is the correct option.

Note:
While solving problems in which data is given in different units, we should convert all the data into one unit. This might be the standard unit or the CGS unit or something else. But, it should be converted into one common unit as it makes our problem easier to calculate and less clumsy.