charge of -3.7nC and charge of -6.95nC, seperated by distance of 40.0cm. Find po
ID: 1705083 • Letter: C
Question
charge of -3.7nC and charge of -6.95nC, seperated by distance of 40.0cm. Find position at which a 3rd charge of+7.9nC can be placed so net electrostatic force = zero.Question (A) ...........Cm form the -3.7nC Charge
Question (B)............Cm fromthe -6.95 nC charge
Prob (2)
Two hard rubber spheres of mass 11 g are rubbed vigorously with fur on a dry day and are then suspended from a rod with two insulating strings of length 5.0 cm. They are observed to hang at equilibrium as shown in the figure below, each at an angle of 10° with the vertical. Calculate the amount of charge that is found on each sphere. (Problem is courtesy of E. F. Redish.)
Explanation / Answer
charges q = -3.7nC = -3.7 * 10 ^ -9 C
q ' = -6.95nC = -6.95 * 10 ^ -9 C
seperation distance r = 40.0cm
3rd charge Q = +7.9nC = 7.9 * 10 ^ -9 C
if the force on 3rd charge due to q is equal and opposite to the force on Q due to q' then net force is zero.
i.e., force on Q due to q = force on Q due to q '
K Qq / x^ 2 = KQq' / ( r-x) ^ 2
where x = distance of the required point from charge q
q / x^ 2 = q ' / ( r-x) ^ 2
sqrt ( q / q' ) = x / ( r- x)
x / (40-x) = 0.7296
x = 29.18 - 0.7296 x
1.7296 x = 29.18
x = 16.87 cm
Question (A) 16.87 Cm form the -3.7nC Charge
Question (B23.12Cm fromthe -6.95 nC charge
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