- Home
- Standard 12
- Physics
2. Electric Potential and Capacitance
medium
Electric potential is given by
$V = 6x - 8x{y^2} - 8y + 6yz - 4{z^2}$
Then electric force acting on $2\,C$ point charge placed on origin will be......$N$
A
$2$
B
$6$
C
$8$
D
$20$
Solution
(d)${E_x} = – \frac{{dV}}{{dx}} = – (6 – 8{y^2}),$${E_y} = – \frac{{dV}}{{dy}} = – ( – \,16xy – 8 + 6z)$
${E_z} = – \frac{{dV}}{{dz}} = – \,(6y – 8z)$
At origin $x = y = z = 0$ so,${E_x} = – \,6,\,{E_y} = 8$ and ${E_z} = 0$
$==>$ $E = \sqrt {E_x^2 + E_y^2} = 10\,N/C$.
Hence force $F = QE = 2 \times 10 = 20\,N$
Standard 12
Physics
