The magnitude of the two charges is doubled and the distance of their separation is also doubled. The electrostatic force between them will :

Question

The magnitude of the two charges is doubled and the distance of their separation is also doubled. The electrostatic force between them will : (A) be halved (B) doubled (C) become four times (D) remain unchanged

Tardigrade Admin · Accepted Answer

F1 = (1/4 π ε0) (q1q2/r2) and F2 = (1/4 π ε0) = ((2q1)(2q2)/(2r)2) = F1

Q. The magnitude of the two charges is doubled and the distance of their separation is also doubled. The electrostatic force between them will :

be halved

doubled

become four times

remain unchanged

$F_{1} = \frac{1}{4 π ε _{0}} \frac{q _{1} q _{2}}{r ^{2}}$ and $F_{2} = \frac{1}{4 π ε _{0}} = \frac{( 2 q _{1} ) ( 2 q _{2} )}{( 2 r ) ^{2}} = F_{1}$

Q. The magnitude of the two charges is doubled and the distance of their separation is also doubled. The electrostatic force between them will :

be halved

doubled

become four times

remain unchanged

F1​=4πε0​1​r2q1​q2​​ and F2​=4πε0​1​=(2r)2(2q1​)(2q2​)​=F1​

$F_{1} = \frac{1}{4 π ε _{0}} \frac{q _{1} q _{2}}{r ^{2}}$ and $F_{2} = \frac{1}{4 π ε _{0}} = \frac{( 2 q _{1} ) ( 2 q _{2} )}{( 2 r ) ^{2}} = F_{1}$