Question

Charges 4Q, q and Q and placed along

x-axis at positions x=0, x=l/2 and x=l,

respectively. Find the value of q so that

force on charge Q is zero​

Answer 1

Answer:

The value of q so that force on Q is zero is q = -Q.

Explanation:

• Coulomb's law states that, "The force of attraction or repulsion between two charged particles is directly proportional to product of magnitude of charges and inversely proportional to square of distance between them."
• Let, q1 and q2 be two charges separate by distance r. Then force between charge is given by

$k \frac{q1q2}{ {r}^{2} }$

where, k is proportionality constant.

Solution:

• Let, charge 4Q be at point A ( x = 0 ) , charge q be at point B ( x = 1/2 ) and charge Q be at point C ( x = 1 ).
• The force of repulsion between charge 4Q and Q is, F1 given by:

$F1 = - k \frac{4 {Q}^{2} }{ {1}^{2} } \\ = - 4k {Q}^{2}$

• negative sign shows repulsive force.
• The force of attraction between charge Q and q is, F2 given by :

$F2= k \frac{Qq}{ { (\frac{1}{2}) }^{2} } \\ = 4kQq$

• The net force on charge Q will be zero if F1 = F2. This gives:

$- 4k {Q}^{2} = 4kQq \\ q = - Q$

• Hence, magnitude of charge q will be equal to Q but opposite in polarity.

Answer 2

Answer:

The value of charge q is -Q.

Explanation:

Given three charges 4Q, q and Q and placed along x-axis.

Positions of the three charges are  $x=0$, $x=l/2$ and $x=l$

According to Coulomb's law, the electrostatic force acting between two charges $q_1$ and $q_2$ separated by a distance $r$ is given by

$F=\dfrac{1}{4\pi \epsilon_{0} } \dfrac{q_1q_2}{r^{2} }$

Substituting the given values,

the, force acting between 4Q and Q is

$F=\dfrac{1}{4\pi \epsilon_{0} } \dfrac{4QQ}{l^{2} }=\dfrac{1}{4\pi \epsilon_{0} } \dfrac{4Q^{2} }{l^{2} }$

And the force acting between q and Q is

$F=\dfrac{1}{4\pi \epsilon_{0} } \dfrac{Qq}{(l/2)^{2} }$

Total force acting on charge q is

$F=\dfrac{1}{4\pi \epsilon_{0} } \dfrac{4Q^{2} }{l^{2} }+\dfrac{1}{4\pi \epsilon_{0} } \dfrac{Qq}{(l/2)^{2} }$

Given that  force on charge Q is zero​, therefore

$F=\dfrac{1}{4\pi \epsilon_{0} } \dfrac{4Q^{2} }{l^{2} }+\dfrac{1}{4\pi \epsilon_{0} } \dfrac{Qq}{(l/2)^{2} }=0$

$\implies \dfrac{4Q^{2} }{l^{2} }+ \dfrac{Qq}{(l/2)^{2} }=0$

$\implies \dfrac{4Q^{2} }{l^{2} }=- \dfrac{4Qq}{l^{2} }$

$\implies Q=- q$ or $q=-Q$

Therefore, the value of charge q is -Q.