The point charges in Figure 17.41 are located at the corners of an equilateral t

Question

The point charges in Figure 17.41 are located at the corners of an equilateral triangle 25.0 cm on a side, where qb = +12.0 µC and qc = -5.50 µC. (Assume that the +x-axis is directed to the right.)

                qa

                  *

*                                  *

qb                                  qc

Find the electric field at the location of qa.
Magnitud/Direction ° (counterclockwise from the +x-axis)

(b) What is the force on qa, given that qa = +1.50 nC?
Magnitude/Direction (counterclockwise from the +x-axis)

Explanation / Answer

Eelectric field due to a charge at distance r is given by

E = kq/ r^2

field due to qb

Eqb = kqb / a^2 = 9 x 10^9 x 12 x 10^-6 / 0.25^2 = 1728000 N/C

direction is along qb -> qa

vector Eqb = 1728000 [ cos60 i + sin60j] = 864000i + 1469492 j N/C

field due to qc,

Eqc = 9 x 10^9 x 5.50 x 10^-6 / 0.25^2 = 792000 N/C

direction is along qa -> qc

vector Eqc = 792000 [ cos60i - sin60 j] = 396000i - 685892j N/C

E = Eqb + Eqc = 1260000i + 783600 j N/C

magnitude = sqrt ( i^2 + j^2 ) = 1483788.72 N/C

direction = tan-1(j/i) = 31.88 degrees

b) F = qE = 1483788.72 x 1.50 x 10^-9 = 2.23 x 10^-3 N

direvtion same yto electric field

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