Question

Three point charges, 9, 4.00 PC, 42 - 2.00 pC, and Q - -1.00 C are located on the vertices of a right triangle as shown in the figure below. The distances d, -40 cm and d-2.0 cm. (a) Calculate the magnitude of the electric field at position of the charge due to the charge ? NC (b) Determine the direction of thelectric field. O toward the left of the screen O toward the night of the screen O toward the top of the screen O toward the bottom of the screen (c) Calculate the magnitude of the electric field at position of the charge due to the charge ? NIC (d) Determine the direction of the electric field. O toward the night of the screen toward the top of the screen toward the bottom of the screen toward the left of the screen

Answer 1

Electric field is given by:

E = kq/r^2

direction of E will be towards -ve charge and away from +ve charge.

a.)

Now for charge q1

E1 = k*q1/r1^2

here, r1 = distance between q1 and Q = 4.00 cm = 0.04 m

q1 = charge on q1 = 4.00 $\mu$ C = 4.00*10^-6 C

then, E1 = (9*10^9)*(4.0*10^-6)/0.04^2

E1 = 2.25*10^7 N/C

b.)

As, q1 is a positive charge.

then, electric field due to charge q1 will be away from charge for point Q, toward bottom of the screen.

c.)

for charge q2

E2 = k*q2/r2^2

here, r2 = distance between q2 and Q = d2 = 2.00 cm = 0.02 m

q2 = charge on q2 = 2.00 $\mu$ C = 2.00*10^-6 C

then, E2 = (9*10^9)*(2.0*10^-6)/0.02^2

E2 = 4.5*10^7 N/C

d.)

As, q2 is also positive charge.

then, Electric field due to charge q2 will be away from charge for point Q, toward the left of the screen.

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