Question

ABC is a right-angle triangle with right angle $A$ and $AB=AC=30\text{cm}$ if the charge on $B$ and $C$ is $4\times {10}^{-3}\mu \text{C}$ then the electric field at $A$ will be

• A. $400\text{N/C}$ parallel to BC
• B. $800\text{N/C}$ parallel to BC
• C. $400\sqrt{2}\text{N/C}$ perpendicular to BC
• D. $800\text{N/C}$ perpendicular to BC
  

Answer 1

The correct option is C $400\sqrt{2}\text{N/C}$ perpendicular to BC


Given:

$q_B=q_C=4\times {10}^{-3}\mu \text{C}$

$AB=AC=r=30\text{cm}$

As we know that the electric field due to point charge is given by, $E=\frac{Kq}{r^2}$

From figure it is clear that,

electric field due to B and C at A is same i.e. $E$

$E=\frac{F}{g}=\frac{K{q}^2}{r^2}=\frac{Kq}{r^2}=\frac{9\times {10}^9\times 4\times {10}^{-3}\times {10}^{-6}}{(30\times {10}^{-2}{)}^2}$

$E=400\text{N/C}$

Now the resultant electric field on A is,

$E_{net}=\sqrt{E^2+E^2+2EE\cos \left({90}^{\circ }\right)}=\sqrt{2}E$

$E_{net}=400\sqrt{2}\text{N/C}$