1
JEE Main 2019 (Online) 9th April Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
Four point charges –q, +q, +q and –q are placed on y-axis at y = –2d, y = –d, y = +d and y = +2d, respectively. The magnitude of the electric field E at a point on the x-axis at x = D, with D >> d, will behave as :-
A
$$E \propto {1 \over D^3}$$
B
$$E \propto {1 \over D}$$
C
$$E \propto {1 \over D^4}$$
D
$$E \propto {1 \over D^2}$$
2
JEE Main 2019 (Online) 9th April Morning Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
A system of three charges are placed as shown in the figure :

JEE Main 2019 (Online) 9th April Morning Slot Physics - Electrostatics Question 155 English If D >> d, the potential energy of the system is best given by :
A
$${1 \over {4\pi {\varepsilon _0}}}\left[ { {{{q^2}} \over d} + {{qQd} \over {{D^2}}}} \right]$$
B
$${1 \over {4\pi {\varepsilon _0}}}\left[ { - {{{q^2}} \over d} - {{qQd} \over {2{D^2}}}} \right]$$
C
$${1 \over {4\pi {\varepsilon _0}}}\left[ { - {{{q^2}} \over d} - {{qQd} \over {{D^2}}}} \right]$$
D
$${1 \over {4\pi {\varepsilon _0}}}\left[ { - {{{q^2}} \over d} + {2{qQd} \over {{D^2}}}} \right]$$
3
JEE Main 2019 (Online) 8th April Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
A positive point charge is released from rest at a distance r0 from a positive line charge with uniform density. The speed (v) of the point charge, as a function of instantaneous distance r from line charge, is proportional to :- JEE Main 2019 (Online) 8th April Evening Slot Physics - Electrostatics Question 157 English
A
$$v \propto \left( {{r \over {{r_0}}}} \right)$$
B
$$v \propto \ln \left( {{r \over {{r_0}}}} \right)$$
C
$$v \propto {e^{ + r/{r_0}}}$$
D
$$v \propto \sqrt {\ln \left( {{r \over {{r_0}}}} \right)} $$
4
JEE Main 2019 (Online) 8th April Evening Slot
MCQ (Single Correct Answer)
+4
-1
Change Language
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 :-
A
–520 V
B
180 V
C
–48 V
D
320 V
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