A point charge of -3 mu C is located at x = 2 m, y = -2 m. A second point charge

Question

A point charge of -3 mu C is located at x = 2 m, y = -2 m. A second point charge of 12 mu C is located at x = 1 m, y = 3 m. (a) Find the magnitude and direction of the electric field at x = -1 m, y = 0. N/C 5 attempt(s) made (maximum allowed for credit = 5) [after that, multiply credit by 0.5 up to 10 attempts] degree (counterclockwise from plus x axis) 0 attempt(s) made (maximum allowed for credit = 5) [after that, multiply credit by 0.5 up to 10 attempts] (b) Calculate the magnitude and direction of the force on an electron at x = -1 m, y = 0. N 0 attempt(s) made (maximum allowed for credit = 5) [after that, multiply credit by 0.5 up to 10 attempts] degree (counterclockwise from plus x axis) 0 attempt(s) made (maximum allowed for credit = 5) [after that, multiply credit by 0.5 up to 10 attempts]

Explanation / Answer

a)

Electric field due to -3uC :

E1 = k*(3*10^-6)/(3^2 + 2^2)

= 2076.9 N

in vector form : E1 = 2076.9*(-3/sqrt(3^2 + 2^2) i + 2/sqrt(3^2 + 2^1) j)

= -1728.1 i + 1152.1 j

Similalrly, for the 12 uC change :

E2 = (9*10^9*12*10^-6/(2^2 + 3^2))*(-2/sqrt(2^2 + 3^2) i + -3/sqrt(2^2+ 3^2) j)

= 8307.7*(-0.555 i - 0.832 j)

= -4610.8 i - 6912 j

So, net electric field = E1 + E2

=  (1728.1 i - 1152.1 j ) + (4610.8 i + 6912 j )

= 6338.9 i + 5759.9 j

So, magnitude of electric field = sqrt(6338.9^2 + 5759.9^2)

= 8564.9 N/C

direction = atan(5759.9/6338.9)

= 42.3 deg

b)

magnitude of force = 8564.9*1.6*10^-19

= 1.37*10^-15 N

direction = 42.3 + 180 = 222.3 deg

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