Question

Two point charges $+4q$ and $+q$ are placed $30cm$ apart. At what point on the line joining them is the electric field is zero?

• A. $5cm$ from charge $4q$
• B. $20cm$ from charge $q$
• C. $5cm$ from charge $q$
• D. $20cm$ from charge $4q$

Answer 1

The correct option is D $20cm$ from charge $4q$

Let net electric field is zero at point $xcm$ from $+q$charge.

ie at a distance $(30-x)cm$ from $+q$ charge.

$E_1=$electric field due to $+q$ charge at point $x$.

$=9\times {10}^9\times \frac{q}{x^2}$

$E_2=$electric field due to $+4q$

charge at point ${}^{\prime }{x}^{\prime }$

ie $(30-x)$

$E_2=9\times {10}^9\times \frac{4q}{{(30-x)}^2}$

Net electric field

$E=E_1-E_2=0ie{E}_1=E_2\frac{4q}{{(30-x)}^2}=\frac{q}{x^2}\Rightarrow 4x^2={(30-x)}^2\Rightarrow 2x=30-x\Rightarrow 3x=30cm\Rightarrow x=10cmfrom+qcharge.ie30-10=20cmfrom+4qcharge.$

so, correct answer is option (d).