C harge q 1 = 7.45 µC is at the origin, and charge q 2 = -4.80 µC is on the x -a

Question

Charge q1 = 7.45 µC is at the origin, and charge q2 = -4.80 µC is on the x-axis, 0.300 m from the origin (see figure).

(a) Find the magnitude and direction of the electric field at point P, which has coordinates (0, 0.400) m.

direction

(c) Place a charge of -6.10 µC at point P and find the magnitude and direction of the electric field at the location of q2 due to q1 = 7.45 µC and the charge at P.

(d) Find the magnitude and direction of the force on q2.

direction

Explanation / Answer

a) @ = tan^-1(0.4/0.3) = 53 deg

magnitude of field strength due to a charge point = kq /r^2

due to q1 :

E1 = (9 x 10^-9 x 7.45 x 10^-6)/(0.4^2)   (j)

E1 = 419062.5 j N/C

due to q2:

E2 =[9 x10^-9x 4.80 x 10^-9 / (0.4^2 + 0.3^2) ] ( cos53i - sin53 j)

E2 = 103993.64i - 138004.22 j N/C

Enet = E1 + E2 = 103993.64i + 281058.28j N/C

magnitude = sqrt(103993.64^2 + 281058.28^2 ) = 299680.56 N/C

direction = tan^-1(281058.28 / 103992.64) = 70 deg

B) F = qE = 2 x 10^-8 x 299680.56 = 6 x 10^-3 N

direction = 70 deg

c) due to q1 :

E1 = (9 x 10^-9 x 7.45 x 10^-6)/(0.3^2)   (j)

E1 = 745000 i   N/C

due to P:

E2 =[9 x10^-9x 6.10 x 10^-9 / (0.4^2 + 0.3^2) ] ( - cos53i + sin53 j)

E2 = - 132158.58i + 175380.36j N/C

Enet = E1 + E2 = - 612841.42 + 175380.36j N/C

magnitude = sqrt(612841.42^2 + 175380.36^2 ) = 637442.45 N/C

direction = tan^-1(175380.36 / 612841.42) = 15.97 deg

d) F = qE

F = 4.80 x 10^-6 x 637442.45 = 3.06 N

direction = 180 + 15.97 = 195.97 deg

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