Question

For the configuration of the two charges shown , the electric field will be zero from charge +Q at distance equals to​

Answer 1

For the configuration of the two charges shown, the electric field will be zero from charge +Q at distance equals to​ d/3

Explanation:

Both the charges are positive charges thus they will repel each other

Electric field will be zero where the net force of repulsion at any positive charge Q will be zero

Let us assume that point is at distance x fro +Q charge

Equating the electrostatic force

$k\frac{Q\times Q}{x^2}=k\frac{Q\times 4Q}{(d-x)^2}$

$\implies \frac{1}{x^2}=\frac{4}{(d-x)^2}$

$\implies (d-x)^2=4x^2$

$\implies d-x=2x$

$\implies x=\frac{d}{3}$

Hope this answer is helpful.

Know More:

Q: Two charges 4 and 36 microcoulomb are placed 60cm apart . At what distance from the larger charge is the electric field intensity 0?

Click Here: https://brainly.in/question/1757742

Answer 2

Given:

Two charges have been shown.

To find:

Distance from +Q at which electric field will be zero.

Calculation:

• The point where electric field is zero is called NEUTRAL POINT.

• Lets assume that the neutral point is "x" distance from +Q, and "(d-x)" from +4Q.

$\therefore \: E1 = E2$

$\implies \dfrac{k(Q)}{ {x}^{2} } = \dfrac{k(4Q)}{ {(d - x)}^{2} }$

• "k" is COULOMB'S CONSTANT.

$\implies \dfrac{1}{ {x}^{2} } = \dfrac{4}{ {(d - x)}^{2} }$

Taking Square Root:

$\implies \dfrac{1}{x} = \dfrac{2}{d - x }$

$\implies d - x = 2x$

$\implies 3x = d$

$\implies x = \dfrac{d}{3}$

So, the NEUTRAL POINT is d/3 distance from +Q charge.