Find the magnitude of the electric field. 006 (part 1 of 2) 10.0 points Three po

Question

Find the magnitude of the electric field.

006 (part 1 of 2) 10.0 points Three positive charges are arranged as shown at the corners of a rectangle. 04 nC 0.612 m 1.52 nC 1.836 m 134 nC Find the magnitude of the electric field at the fourth corner of the rectangle. The value of the Coulomb constant is 8.98755 × 109 N·m2/C2. Answer in units of N/C. 007 (part 2 of 2) 10.0 points What is the direction of this electric field (as an angle measured from the positive:r axis, within limits of -180 to 180°, with counterclockwise positive)? Answer in units of

Explanation / Answer

apply the formula for Electric field due to each of the charge as E = KQ/r^2

so

here at fourth corner,

two fields E1 (due to 1.52 nC) and E2 due to (3.04 nC) acts at right angles

so E1 = 9e9 * 1.52 e-9/(0.612*0.612)

E1 = 36.52 N/C

E2 = 9e9 * 3.04 e -9/(1.836*1.836)

E2 = 8.11 NC

Now Enet Due to these two is Enet^2 = E1^2 +E2^2

E12^2 = 36.52 ^2 + 8.11^2

E12^2 = 1399.588

E12 = 37.41 N/C

also between E1 and E2 is tan theta = 36.52/8.11

theta = 77.48deg

now for E3, distance d^2 = 0.612^2 + 1.836^2

d^2 = 3.745

so

E3 = (9e9 * 1.34e-9/(3.745)

E3 = 3.22 N/C this points at 45 deg

net angle between E3 and E12 os 77.48 -45 = 32.47 deg

so

Enet due to at the three is

ENet^2 = E12^2 E3^2 +2E12E3cos theta

Enet ^2   = 37.41^2 + 3.22^2 + (2* 3.22 * 37.41 * cos 32.47)

Enet = 40.16 N/C

Direction tan theta = 37.41/3.22

theta = 85 deg + ve x axis

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