
Two balls carrying charges $ + 7\;{\rm{\mu C}}$ and $ - 5\;{\rm{\mu C}}$ attract each other with a force $F$. If a charge $ - 2\;{\rm{\mu C}}$ is added to both, the force between them will be: -
(1) $F$
(2) $\dfrac{F}{2}$
(3) $2F$
(4) Zero
Answer
570.6k+ views
Hint:Here, we will use the Coulomb's formula for attraction or repulsion force between two charges. If two charges have opposite signs then the force acting between them is attractive and if the two charges have the same sign then the force acting between the two charges is considered as repulsive force.
Complete step by step answer:
Given: The initial value of first charge is ${q_1} = + 7\;{\rm{\mu C}}$ and initial value of second charge is ${q_2} = - 5\;{\rm{\mu C}}$. The force acting between the two charges is $F$. The new charge added to both charges is $q = - 2\;{\rm{\mu C}}$.
The formula for attraction or repulsion force between two charges is ${F_1} = k\dfrac{{{q_1}{q_2}}}{{{r^2}}}$
Here, k is Coulomb's constant and r is the distance between the two charges.
Substituting the initial values of charge in above formula.
$\begin{array}{l}
F = {F_1} = k\dfrac{{7 \times \left( { - 5} \right)}}{{{r^2}}}\\
{F_1} = - 35\dfrac{k}{{{r^2}}}
\end{array}$
Now, a new charge is added to the initial values of first and second charge. Then, the obtained new values of first and second charges are,
$\begin{array}{l}
{q_3} = 7 - 2\\
{q_3} = 5\;{\rm{\mu C}}
\end{array}$ and $\begin{array}{l}
{q_4} = - 5 - 2\\
{q_4} = - 7\;{\rm{\mu C}}
\end{array}$
Now, we substitute the obtained new value of both charges in Coulomb's formula to get the force between two charges,
$\begin{array}{l}
{F_2} = k\dfrac{{{q_3}{q_4}}}{{{r^2}}}\\
{F_2} = k\dfrac{{5 \times \left( { - 7} \right)}}{{{r^2}}}\\
{F_2} = - 35\dfrac{k}{{{r^2}}}
\end{array}$
We can see that both the values of attraction force are the same and are equal to the initial value of force.
$F = {F_1} = {F_2}$
Therefore, the force between them will be $F$ .
So, the correct answer is “Option 1”.
Note:
The attraction or repulsion force acting between two charges of different values is inversely proportional to the distance between the two charges. A new charge is added to the initial values of first and second charge.
Complete step by step answer:
Given: The initial value of first charge is ${q_1} = + 7\;{\rm{\mu C}}$ and initial value of second charge is ${q_2} = - 5\;{\rm{\mu C}}$. The force acting between the two charges is $F$. The new charge added to both charges is $q = - 2\;{\rm{\mu C}}$.
The formula for attraction or repulsion force between two charges is ${F_1} = k\dfrac{{{q_1}{q_2}}}{{{r^2}}}$
Here, k is Coulomb's constant and r is the distance between the two charges.
Substituting the initial values of charge in above formula.
$\begin{array}{l}
F = {F_1} = k\dfrac{{7 \times \left( { - 5} \right)}}{{{r^2}}}\\
{F_1} = - 35\dfrac{k}{{{r^2}}}
\end{array}$
Now, a new charge is added to the initial values of first and second charge. Then, the obtained new values of first and second charges are,
$\begin{array}{l}
{q_3} = 7 - 2\\
{q_3} = 5\;{\rm{\mu C}}
\end{array}$ and $\begin{array}{l}
{q_4} = - 5 - 2\\
{q_4} = - 7\;{\rm{\mu C}}
\end{array}$
Now, we substitute the obtained new value of both charges in Coulomb's formula to get the force between two charges,
$\begin{array}{l}
{F_2} = k\dfrac{{{q_3}{q_4}}}{{{r^2}}}\\
{F_2} = k\dfrac{{5 \times \left( { - 7} \right)}}{{{r^2}}}\\
{F_2} = - 35\dfrac{k}{{{r^2}}}
\end{array}$
We can see that both the values of attraction force are the same and are equal to the initial value of force.
$F = {F_1} = {F_2}$
Therefore, the force between them will be $F$ .
So, the correct answer is “Option 1”.
Note:
The attraction or repulsion force acting between two charges of different values is inversely proportional to the distance between the two charges. A new charge is added to the initial values of first and second charge.
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