Question

Related to the diagram of two charges, $+Q$ and $-4Q$. At which point does the net electric field vector point to the left?

• A. $A$
• B. $B$
• C. $C$
• D. $D$

Answer 1

The correct option is D $D$

Assuming Unit distance between each markings on scale.

$E=\frac{k\times q}{r^2}$

Negative charge attracts and Positive charge repels.

Taking care of Direction:

For A:

$E_A=-\frac{k\times Q}{36}+\frac{k\times \left(4Q\right)}{144}=0$

For B:

$E_B=\frac{k\times Q}{4}+\frac{k\times \left(4Q\right)}{16}=0$

For C:

$E_C=\frac{k\times Q}{16}+\frac{k\times \left(4Q\right)}{4}=\frac{17k\times Q}{16}$

Right Side

For D:

$E_D=\frac{k\times Q}{144}+-\frac{k\times \left(4Q\right)}{36}=-ve$

Left Side

Hence Answer is D