+3q +q -2q +9 +9 +5 +3q (c23p64_6e) In the figure, four charges, given in multip

Question

+3q +q -2q +9 +9 +5 +3q (c23p64_6e) In the figure, four charges, given in multiples of 3.00x10-6 C form the corners of a square and four more charges lie at the midpoints of the sides of the square. The distance between adjacent charges on the perimeter of the square is d - 7.00x10-3 m. What are the magnitude and direction of the electric field at the center of the square? The magnitude of E? 1.79e9 N/C Incorrect. Tries 3/8 Previous Tries Submit Answer Ex? 1.51e9 N/OC Submit Answer Incorrect. Tries 1/8 Previous Tries Ey? 9.63e8 N/C Submit Answer Incorrect. Tries 1/8 Previous Tries

Explanation / Answer

field due to a point charge at distance r = kq/r^2

direction -> radially away from positive charge and opposite for negative charge.

distance of centre from corner = sqrt(d^2 + d^2) = 1.414d

field at centre = due to 3q + due to upper q + due to -5q + due to -2q + due to 3q + due to lowerq

+ due to 5q + due to -q

= [ (k(3q)/2d^2) (cos45i - sin45j) ] + [ (k(q)/d^2) (-j)] + [(k(5q)/2d^2) (cos45i + sin45j)] + [(k(2q)/d^2) (i)]

+[(k(3q)/2d^2) (-cos45i + sin45j)] + [(k(q)/d^2) (j)] + [(k(5q)/2d^2) (cos45i + sin45j)] + [(k(q)/d^2) (-i)]

= 2[(k(5q)/2d^2) (cos45i + sin45j)] +[(k(q)/d^2) (i)]

= 4.54kq/d i + 3.54kq/d j

Ex = 4.54 x 9x 10^9 x 3 x 10^-6 / ( 7 x 10^-3)^2 = 2.50 x 10^9 N/C

Ey = 3.54 x 9x 10^9 x 3 x 10^-6 / ( 7 x 10^-3)^2 = 1.95 x 10^9 N/C

E = sqrt(Ex^2 + Ey^2) = 3.17 x 10^9 N/C

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