(refer to photos attached. Example of previous question with wrong/correct answers example, and current question needing to be solved)

Determine the electric field strength at a point 1.00 cm to the left of the middle charge shown in the figure below. (Enter the magnitude of the electric field only.) _____N/C

If a charge of −3.94 µC is placed at this point, what are the magnitude and direction of the force on it?

Magnitude _______N

Direction?

- toward the left
- upward
-downward
- toward the right


Refer To Photos Attached Example Of Previous Question With Wrongcorrect Answers Example And Current Question Needing To Be Solved Determine The Electric Field S class=
Refer To Photos Attached Example Of Previous Question With Wrongcorrect Answers Example And Current Question Needing To Be Solved Determine The Electric Field S class=

Sagot :

(a) The electric field strength at a point 1.00 cm to the left of the middle is  2.0 x 10⁷ N/C.

(b) The magnitude of the force is 94.4 N and direction of the force on it towards the left.

Electric field strength

The electric field strength at a point 1.00 cm to the left of the middle is calculated as follows;

E = kq/r²

Electric field due to first charge

E1 = (9 x 10⁹ x 6 x 10⁻⁶)/(0.02)²

E1 = 1.35 x 10⁸ N/C

Electric field due to second charge

E2 =  -(9 x 10⁹ x 1.5 x 10⁻⁶)/(0.01)²

E2 = - 1.35 x 10⁸ N/C

Electric field due to third charge

E3 = - (9 x 10⁹ x 2 x 10⁻⁶)/(0.03)²

E3 = -2.0 x 10⁷ N/C

Net electric field

E = E1 + E2 + E3

E = +1.35 x 10⁸ N/C - 1.35 x 10⁸ N/C - (-2.0 x 10⁷ N/C)

E = +2.0 x 10⁷ N/C

Force on the charge −4.72 µC

F = Eq

F = 2.0 x 10⁷ x -4.72 x 10⁻⁶

F = -94.4 N

Thus, the direction of the force will be towards the left.

Learn more about force on charge here: brainly.com/question/25923373

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