Question

Two point charges +q and +4q are separated by a distance 6a. Find the point joining the two charges where electric field is zero.

Answer 1

Here is your ans.

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Answer 2

Answer:

2a from charge q and 4a from charge 4q

Explanation:

Total distance is $6a$

Let the distance at which the electric field is zero be $x$ from $+q$ charge.

Now, for Electric Field $(B)$ to be $0$, the Coulomb forces should cancel each other out.

For this we can assume a charge $q_{0}$ placed between the charges such that net force on it due to both charges is 0 i.e. $F_{q-0q}  = F_{4q-q0}$

( We consider only magnitudes here, Since If we consider negative sign it will lead to a wrong answer..... Starting with vectors will fetch us the same relation of positive signs )

Therefore, We know Coulombian force,

$F = k\frac{q1 q2}{r^{2} }$

$k\frac{q q_{0} }{x^2} = k\frac{q_{0} 4q}{(6a-x)^2}$

$\frac{1 }{x^2}=\frac{4}{(6a-x)^2}$

Taking square roots on both sides. PLEASE CONSIDER ONLY POSITIVE VALUES.

Hence,

$\frac{1}{x} =\frac{2}{6a-x}$

${6a-x}={2x}$$6a=3x$

$2a=x$

The required answer or distance is '$2a$' from charge $q$ or '$4a$' from charge $4q$.

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