Electric Charges Fields And Gauss Law(11th And 12th > Physics ) Questions and Answers
Explanation:-
Answer: Option C. -> Along the diagonal BD:
C
We put a unit positive charge at O. Resultant force due to the charge placed at A and C is zero and resultant charge due to B and D is towards D along the diagonal BD.
Six charges, three positive and three negative of equal magnitude are to be placed at the vertices of a regular hexagon such that the electric field at O is double the electric field when only one positive charge of same magnitude is placed at R. Which of the following arrangements of charges is possible for P, Q, R, S, T and U respectively
Explanation:-
Answer: Option D. -> -,+,+,-,+,-:
D
If the charges are arranged according to the option (d), the electric fields due to P and S and due to Q and T add to zero, while due to U and R will be added up.
Explanation:-
Answer: Option C. -> 10√2:
C
Body moves along the parabolic path.
For vertical motion : By using v=u+at
⇒ vy=0+QEm.t=10−6×10310−3×10=10m/sec
For horizontal motion - it;s horizontal velocity remains the same i.e after the same i.e after 10 sec, horizontal
velocity of body vx=10 m/sec.
Velocity aftey 10 sec v=√V2x+V2y=10√2m/sec
Explanation:-
Answer: Option C. -> The magnitude of velocity of the electron will first decrease and then increase:
C
As electric field applies the force on electron in the direction opposite to it’s motion, the velocity of the electron will decrease.
Explanation:-
Answer: Option D. -> If the assertion and reason both are false.:
D
Gravitational force is the dominating force in nature and not coulomb's force. Gravitational force is the weakest force. Also, Coulomb's force>> gravitational force.
Explanation:-
Answer: Option A. -> -q:
A
The force between 4q and q; F1=14πϵ0.4q×ql2
The force between Q and q; F2=14πϵ0.Q×q(l2)2
We want F1+F2=0 or 4q2l2=−4Qql2⇒Q=−q