Two point charges, Q_1 = -40 R^C and Q_2 = +50 mu C are separated by a distance
ID: 1644107 • Letter: T
Question
Two point charges, Q_1 = -40 R^C and Q_2 = +50 mu C are separated by a distance of 12 cm. The electric field at the point P is zero. How far from Q_1 is P? Particles of charge Q_1 = +5.00 mu C. Q_2 = -6.00 mu C and Q_3 = +8.00 mu C are placed on the corners of a square of side 0.400 m as shown below. Calculate the force an Q_2 (Magnitude and direction). A +50 mu C point charge is separated from a -25 mu C charge by a distance of 0.25 m as shown below. (A) First calculate the electric field at midway between the two charges. (B) Find the force on an electron that is placed at this point and (C) Calculate the acceleration when it is released. Calculate the electric potential of the upper right corner of a rectangle that is 5.0 cm high and 8.0 cm long if there are charges of 5 mu C m the upper left corner. +2 mu C in the lower left corner and -6 mu C in the lower right corner Assume zero is at infinity. A point charge Q creates an electric potential of -165 V at a distance of 15 cm. What is Q? An electronic is build to store energy in 500 mu F at 110 V (a) How much electric energy can be stored? (b) What is the power release if nearly all this energy is released in 5 ?Explanation / Answer
(6)
formula for electric field due to point charge is
E = kq/r^2
electric field due to Q1 is
E1 = k (-40 uC)/ x^2
electric field due to 50 uC is
E2 = k ( 50 uC)/ ( 12 cm + x)^2
at point P net electric field is zero
E1+E2 = 0
k (-40 uC)/ x^2+k ( 50 uC)/ ( 12 cm + x)^2=0
k ( 50 uC)/ ( 12 cm + x)^2= k (40 uC)/ x^2
50/40 = ( 12+x)^2/x^2
12+x/x = sqrt 50/40
12+x= 1.11x
12 = 0.11 x
x= 12/0.11
=109.09 cm
(8)
(a)
electric field at mid point is
E net = E1 + E2
= 9*10^9 (50 * 10^-6)/(0.25/2)^2 + 9*10^9 (25 * 10^-6)/(0.25/2)^2
=4.32 * 10^7 N/C
F = Enet q = 4.32* 10^7 N/C ( 1.6* 10^-19) = 6.91 * 10^-12 N
a= F/m = 6.91 * 10^-12 N/9.11* 10^-31 = 7.58 * 10^18 m/s^2
10)
V = kq/r
-165= 9* 10^9 q/0.15
q= -2.75 * 10^-9 C
11)
U = 1/2 C V^2
= 0.5 ( 500* 10^-6) ( 110)^2
=3.025 J
P = U/t = 3.025 J/5 s = 0.605 W
as per guide lines I worked four problems , please post remaining questions in the next post
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