Version f PHY-112 Practice Midterm -Spring 2016 Multiple Choice Identily the cho

Question

Version f PHY-112 Practice Midterm -Spring 2016 Multiple Choice Identily the choice that best completes the statement or answers the question 1. A charge of+2Q is placed on an isolated metal sphere and a charge of 40 is placed on an identieat isofeted metal sphere. If the two spheres are briefly connected with a conducting wire and then diseonneeted, what io the charge on the first sphere? E. +30 B. -20 C. +20 A-5.2 pC charge is placed at x-o and a -3.5 third charge be placed so it would be in equilibrium? A. 0.50 m

Explanation / Answer

1. ans: option D
when connected with wire , charges on both the spheres are redistributed uniformly on the combined system ie +2Q -4Q = -2Q on two spheres, when separated, each sphere carries equal charge of - Q .
2 ans : option A
As both charges are negative, , they will have the same effect on the third charge , ie they both will repel in case of negative 3rd charge and both will attract in case of positive 3rd charge.
For equilibrium, charge has to be between these charges , and closer to the smaller chareg.

3. ans: option A
attarction force F = k.q1.q2 / r^2
when charges are halved q1 charge becomes q1 /2 similarly q2 charge becomes q2 /2
F_new = k . (q1 /2).(q2 /2)/r^2 = F/4 , ie force is quatered.
4. ans :option E
A closer look on figure reveals that, charge A has more field lines than charge B , as more field lines means higher charge, so the magnitude of charge B is lesser than 6uC.
Also field lines of A and B are directeed away from each other , ie they are repelling each other , in other words they are same signed chrges
. this makes -4uC as answer.
5. ans :option
Work done = KE = F.d
440 eV = F.0.25
F = 1760 e
F = q.E
1760 e = e.E
E =1760N/c

6. ans: option B
Force F = m . a
In an electric field, F = q . E
For an electron in field E,
q . E = m . a
=> a = qE/m , hence mass is also required here.

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