1) The ratio of electrostatic to gravitational forces between two electrons can be expressed as the ratio of the electrostatic force constant ke2 to the gravitational force constant Gm2. 2) A charged oil droplet of +8.45x10-19 C is found to remain suspended in an electric field of 5.92x104 N/C directed downward. The charge on the droplet is calculated to be -4.85x10-19 C. 3) For a uniform electric field oriented in the x-direction through a cube, the net outward electric flux through the sides of the cube is zero, as the flux in the x-direction is canceled by an equal flux in the opposite

1. 7. The ratio of electrostatic (F.) and gravitational (Fo) forces between two clectrons in vacuum
may be expressed as (b) Fe/Fo ke 4Gm, (a) Fe/FGke/Gm, (c) Fe / FG = 2ke/Grn,- where the
symbols carry their usual meanings. 8. You establish an electric field of magnitude 5.92x 10 N/C
pointing straight down. A charged oil droplet of (d) + 8.45 x 10-19 C mass 2.93 x 10 1s kg is
found to remain suspended against gravity. Find the charge carried by the droplet. (a) + 4.85 x 10
19 C (b) -4.85 x 10-19 C (c)-8.45 x 10 19 C Co Doc Eng 9. Consider a uniform electric field E
oriented in x-direction as shown in Fig. 3. Therefore the net outward electric flux through all
sides of the cube, with each side having a length L, is (b) EL (c) EL/E (a) EL /Eo to being the
permittivity of free space. (d) zero
Solution
7. F e = (k x e 2
)/r 2
F G = (G x m e x m e )/r 2
Therefore, F e /F G = ke 2 /Gm 2
e hence the answer will be (d)
8. Let the charge in the droplet be q. Then, force due to electric field should be equal to the force
due to gravitation for the droplet to be suspended.
So, q x 5.92 x 10 4
= 2.93 x 10 - 15
x 9.8
q = 4.85 x 10 -19
C. Also to get a pull against the grvity the direction of force due to Electric filed
should be upward(opposite direction). Hence the charge on the drop would be negative. Hence
the effective charge on the dropelet would be answer (b) -4.85 x 10 -19 C.
9. Electric flux can be calculated by integrating the electeric field across the area of cross section
in the direction of the area vector. As the electric field is oriented in x-direction, hence the
electric flux along the y-direction and z-direction will be zero(as electric field is perpendicular to
area).
Electric field in the x-direction will be: -EL 2
and EL 2
from the two opposite faces with area
vection in x-direction and x'-direction respectively. Hence the total outward flus will be EL 2
-
EL 2
= 0

2. hence answer will be (d) Zero