Question

Electric potential is given by $V=6x-8x{y}^2-8y+6yz-4z^2$. Then electric force acting on a 2C point charge placed at origin will be?

• A. 2 N
• B. 6 N
• C. 8 N
• D. 20 N

Answer 1

The correct option is D 20 N

Step 1, Given data

Potential is given as $V=6x-8x{y}^2-8y+6yz-4z^2$

Step 2, Finding the electric force at the origin

We know,

$E_x=\frac{-dV}{dx}=-(6-8y^2)$

$E_y=\frac{-dV}{dy}=-(16xy-8+6z)$

$E_z=\frac{-dV}{dz}=-(6y-8z)$

Now we have to find the force at origin

So,

At the origin x=y=z=0

So, $E_x=-6,E_y=8and{E}_z=0$

So now resultant electric field will be

$E=\sqrt{{E_x}^2+{E_y}^2}=\sqrt{{-6}^2+8^2}=10$ N/C
Now,

Force = $Q\times E=2\times 10=20N$

Hence the resultant force is 20 N