JEE MAIN - Physics (2019 - 8th April Evening Slot - No. 19)
The electric field in a region is given by
$$\mathop E\limits^ \to = \left( {Ax + B} \right)\mathop i\limits^ \wedge $$
, where E is in NC–1 and x is in
metres. The values of constants are
A = 20 SI unit and B = 10 SI unit. If the potential
at x = 1 is V1 and that at x = –5 is V2, then
V1 – V2 is :-
–520 V
180 V
–48 V
320 V
Explanation
$$\overrightarrow E = (20x + 10)\widehat i$$
$${V_1} - {V_2} = - \int\limits_{ - 5}^1 {\left( {20x + 10} \right)dx} $$
$${V_1} - {V_2} = - \left( {10{x^2} + 10x} \right)_{ - 5}^1$$
$${V_1} - {V_2} = 10\left( {25 - 5 - 1 - 1} \right)$$
$${V_1} - {V_2} = 180\,V$$
$${V_1} - {V_2} = - \int\limits_{ - 5}^1 {\left( {20x + 10} \right)dx} $$
$${V_1} - {V_2} = - \left( {10{x^2} + 10x} \right)_{ - 5}^1$$
$${V_1} - {V_2} = 10\left( {25 - 5 - 1 - 1} \right)$$
$${V_1} - {V_2} = 180\,V$$
Comments (0)
