Question

Please answer me .last answer was there just write the steps.

Answer 1

let q charge is placed at a distance X from 4q charge between the two charges,then is obviously it is (d-x) apart from q . now for q to be in equilibrium ,the forces by 4q and q must be equal and opposite in direction,as shown in fig-now

forces on q are equal and opposite

so,

here k= 1/4π€°

Answer 2

$\textsf{\underline {\Large{Electric Fields and Charges}}}$ :

$\textsf{\underline {\large{Solution}}}$ :

Let $\mathsf{q_1\:=\:q}$

$\mathsf{q_2\:=\:4q}$

$\mathsf{q_3\:=\:?}$

Let ' x ' be the position where the third charge will be placed.

To be in equilibrium, the charge on third point charge must be negative to balance the other two charges.

The Force exerted by q and 4q point charges will be equal as per the condition of being in equilibrium.

By $\textsf{\underline {\large{Coulomb's Law,}}}$

$\boxed{\mathsf{F\:=\:k\:{\dfrac{q_1q_2}{{r}^{2}}}}}$

$\mathsf{F_{13}\:=\:F_{23}}$

$\mathsf{k\:{\dfrac{q_1q_3}{{d}^{2}}}}}$ = $\mathsf{k\:{\dfrac{q_2q_3}{{d}^{2}}}}}$

$\mathsf{\dfrac{qq_3}{{(\:d\:-\:x\:)}^{2}}}$ = $\mathsf{\dfrac{4qq_3}{{x}^{2}}}$

Taking square root both sides,

$\mathsf{\sqrt{\dfrac{qq_3}{{(\:d\:-\:x\:)}^{2}}}}$ = $\mathsf{\sqrt{\dfrac{4qq_3}{{x}^{2}}}}$

$\mathsf{\dfrac{\sqrt{q}}{(\:d\:-\:x\:)}}$ = $\mathsf{\dfrac{2{\sqrt{q}}}{x}}$

$\mathsf{\dfrac{1}{(\:d\:-\:x}}$ = $\mathsf{\dfrac{2}{x}}$

$\mathsf{x\:=\:2\:(\:d\:-\:x\:)}$

$\mathsf{x\:=\:2d\:-\:2x\:}$

$\mathsf{3x\:=\:2d}$

$\boxed{\mathsf{x\:=\:{\dfrac{2d}{3}}}}$

$\textsf{\underline {\large{Coulomb's Law}}}$ :

It states that the electrostatic force of attraction or repulsion between two stationary point charges is directly proportional to the product of their magnitude and inversely proportional to the square of the distance between them.

⚫This force acts along the line joining between the two charges.

$\boxed{\mathsf{F\:=\:k\:{\dfrac{q_1q_2}{{r}^{2}}}}}$

➡️ While writing the value of charges in the above formula, we do not consider the signs.