My question is 3 parts. Please be show all the work so I can learn how these pro

Question

My question is 3 parts. Please be show all the work so I can learn how these problems are done.

(1)A +13.86 nC charge is located at (0,10.55) cm and a -3.8 nC charge is located (4.43, 0) cm.Where would a -13.6 nC charge need to be located in order that the electric field at the origin be zero?

(2)Moving a charge from point A, where the potential is 398.3489 V, to point B, where the potential is 174.4013 V, takes 0.0005 J of work.  What is the value of the charge?

(3)Two drops of mercury each has a charge on 0.0000000023 nC and a voltage of 243.81 V. If the two drops are merged into one drop, what is the voltage on this drop?

Explanation / Answer

1) Electric fields due to +13.86 nC and -3.8 nC = Electric field due to -13.6 nC

[(-9*10^9*13.86*10^-9)/(0.1055^2)]j + [(9*10^9*3.8*10^-9)/(0.443^2)]i = Electric field due to -13.6 nC = E

-11207.3 j + 174.3 i = Ex i + Ey j

Ex = k*q/x^2 = 174.3

x = sqrt(-9*10^9*13.6*10^-9/174.3) = -0.837 m = -83.7 cm

Ey = -11207.3

k*q/y^2 = -11207.3

(-9*10^9*13.6*10^-9)/(-11207.3) = y^2

y = 0.104 m = 10.4 cm

(x,y) = (-83.7,10.4) cm

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2) W = q*(VB-VA)

charge q = W/(VB-VA)= (0.0005)/(174.4013-398.3489) = -2.23*10^-6 C

3) potential on the new drop is V = n^(2/3)*v

V = 2^(2/3)*243.81 = 387.02 V

n is the number of drops = 2

v is potential of each drop = 243.81 V

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