Point charges are arranged on the vertices of a square with sides of 3.45 cm. St

Question

Point charges are arranged on the vertices of a square with sides of 3.45 cm. Starting at the upper left corner and going clockwise, we have charge A with a check charge of 0.223 mu C, B with a charge of -0.221 mu C, C with a charge of 0.298 mu C, and D with a mass of 3.0 g, but with an unknown charge. Charges A, B, and C are fixed in place, and D is free to move. Particle D's instantaneous acceleration at point D is 293 m/s^2 in a direction theta below the negative x-axis. What is the charge on D? q = mu C

Explanation / Answer

given that

Qa = 0.223 uC

Qb = -0.221 uC

Qc = 0.298 uC

a = 293 m/s^2

m = 3 g

r = 3.45 cm

distance b/w D and B = sqrt(3.45^2 +3.45^2) = 4.87 cm

x component of electric field at D is

Ex = Eax + Ebx + Ecx

Ex = 0 + Eb*cos45 + Ecx

Ex = 9*109*0.221 *10-6 / (0.0487)2 * sqrt2 + 9*109*0.298*10-6 / (0.0345)2

Ex = 2.84*106 N/C

y component of electric field at D

Ey = Eay + Eby+ Ecy

Ey = Eay + Eb*sin45 + 0

Ey = - 9*109*0.223 *10-6 / (0.0345)2 + 9*109*0.221*10-6 / (0.0487)2 *sqrt2

Ey = -1.09*106 N/C

magnitude of electric field

E = sqrt (Ex2 + Ey2)

E = sqrt[ (2.84*106)2 + (1.09*106)2 ]

E = 3.03*106 N/C

we know

F =m*a

also,

F = q*E

so

q = m*a/E

q = 0.003 *293 / 3.03*106

q = 0.290*10-6 C

change on D is 0.290 uC.

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