Name Read each question carefully and provide your answer choice on the answer s

Question

Name Read each question carefully and provide your answer choice on the answer sheet provided. Good Luck! if the result of your calculations for a quantity has SI units of C2 s2/(kg m2), that quantity could be an electric field strength. an electric potential energy an electric potential difference a capacitance a dielectric constant two identical small charged spheres are a certain distance apart, and each one initially experiences an electrostatic force of magnitude F due to the other With time, charge gradually diminishes on both spheres by leaking off When each of the spheres has lost half its initial charge, what will be the magnitude of the electrostatic force on each one? 1/4 F 1/8 F 1/16 F 1/2 F 1/2 F What is the charge on 1 0 kg of protons'(e - 1.60 times 10-9 C,m proton - 1.67 times 10-27 kg) 10CB) 60 times 10 23 9.6-107C 1000C 6.0 times 1026c The electronic potential at a distance of 4 m from a certain point charge is 200 V relative to infinity. What is the potential relative to infinity) at a distance of 2 m from the same charge? 50 V 100 V 400 V 600V Two stationary point charges and 1 are shown in the figure along with a sketch of some field lines representing the electric filed produce by them. What can you deduce from the sketch Q1 and q2 have the same sign; the magnitudes are equal. Q1 is negative and q2 is positive; the magnitude of q1 is less then the magnitude of q2 q1 and q2 have the same sing the magnitude of q1 is greater than the magnitude of q2 q1 is negative and q2 is positive; the magnitude are equal q1 is positive and q2 is negative; the magnitude of q1 is greater than the magnitude of q2.

Explanation / Answer

1) (D) a capacitance

2) (A)F/4
F=kq2/r2 ; Fnew=(k/r2).(q/2).(q/2) =F/4

3) (C) 9.6 x 107C
charge= (1/mproton)xe = 9.6 x 107?C

4) (D) 400V
given, kq/4=200V so kq/2=200x2=400V

5) (B)
because electric field lines converge to a negative charge and go out from positive charge, more number of field lines implies more charge

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