Compute the electric field at a point midway (a) between charges q_2 and q_3 in

Question

Compute the electric field at a point midway (a) between charges q_2 and q_3 in the figure shown below. Use the following diagram for the next 5 questions. What is the distance midway point between q_2 and q_3? a, 100cm b. 60cm c. 20cm d. 0.1m What is the electric field at point A due to q? a. 4.8 times 10 N/C b. 4.2 times 10 N/C c. 3.6 times 10 N/C d. 3.0 times 10 N/C What is the distance between q_1 and the midway point between q_2 and q_3? a. 23.4 cm b. 17.3 cm c. 14.4 cm d. 10cm What is the approximate electric field at point a due to q_1? a. 1.2028467 times 10^6 N/C b. 2.2020000 times 10^6 N/C c. 3.6 times 10^6 N/C d. 4.8 times 10^6 N/C What is the total electric field in the x direction at point a? a. 9.6 times 10^6 N/C b. 8.4 times 10^6 N/C c. 7.2 times 10^6 N/C d. 6.0 times 10^6 N/C

Explanation / Answer

6.

distance between q2 and q3 = 20/2 = 10 cm = 0.1 m

Correct option is D.

7.

electric field is given by:

E = k*q/r^2

Using given values:

E = 9*10^9*4*10^-6/(0.1^2)

E = 3.6*10^6 N/C

Correct option is C.

8.

height of right angle triangle is given by:

h = sqrt (hypotenuse^2 - base^2)

distance = sqrt (20^2 - 10^2) = 17.32 cm

Correct option is B.

9.

E = 9*10^9*4*10^-6/(0.1732^2)

E = 1.20*10^6 N/C

Correct option is A.

10.

Since direction of electric field due to -ve charge is towards the charge and due to +ve charge is away from positive charge, So

Ex = E1 + E2

Since q1 = q2 and r1 = r2, So

E1 = E2

Ex = 3.6*10^6 + 3.6*10^6 = 7.2*10^6

Correct option is C.

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