Question

The electric potential in a certain region is expressed by v = 6x-8xy${}^2$-8y+6yz-4z${}^2$ volts. The magnitude of the force acting on a charge of 2 C situated at the origin will be

• A. 2 N
• B. 6 N
• C. 8 N
• D. 20 N

Answer 1

The correct option is D 20 N

Given expression for electric potential,
v=6x - 8xy${}^2$ - 8y + 6yz - 4z${}^2$ volts
charge Q=2 C
To find, Force (F).
We know,
F=QE...(Formula)
Where Q is charge and E is electric field,
We will solve given expression.
Along X-axis
$=\frac{dv}{dx}=6-8y^2...\left(1\right)$
Along Y-axis
$=\frac{dv}{dy}=-16xy-8+6z...\left(2\right)$
Along Z-axis
$=\frac{dv}{dz}=6y-8z...\left(3\right)$
At origin, from (1), (2) and (3)
$(\frac{dv}{dx},\frac{dv}{dy},\frac{dv}{dz})=(6,-8,0)$
Net electric field,
$E=\sqrt{6^2+(-8{)}^2+(0{)}^2}=\sqrt{100}$
$\therefore E=10N/C$
$\therefore$ F= 2 x 10
$\therefore$F=20 N
Option (D) is correct