Answer : The correct option is, (d) 4 times
Solution :
According to the Coulomb's law, the electrostatic force of attraction or repulsion between two charges is directly proportional to the product of the charges and is inversely proportional to the square of the distance between the the charges.
Formula used :
\(F=k_e\frac{q_1q_2}{r^2}\)
where,
F = electrostatic force of attraction or repulsion
\(k_e\) = Coulomb's constant
\(q_1\) and \(q_2\) are the charges
r = distance between two charges
First we have to calculate the force exerted between S and q when the distance between the charge is 1 unit and let us assumed that the charge be 'q'
\(F_{sq}=k_e\frac{qq}{1^2}=k_e\times q^2\) ..........(1)
Now we have to calculate the force exerted between S and p when the distance between the charge is 2 unit at the same charge.
\(F_{sp}=k_e\frac{qq}{2^2}=k_e\frac{q^2}{4}\) ...........(2)
Equation equation 1 and 2, we get
\(\frac{F_{sq}}{F_{sp}}=\frac{1}{4}\)
\(F_{sq}=4\times F_{sp}\)
Therefore, the force exerted between S and q is 4 times the force exerted between S and p.
