Two point charges are lying on the y axis as in the figure: q 1 = 2.80 µC and q

Question

Two point charges are lying on the y axis as in the figure: q1 = 2.80 µC and q2 = +2.80 µC. They are equidistant from the point P, which lies on the x axis.

Concepts:
(i) There is no charge at P in part a of the figure. Is there an electric field at P?

Yes or No

(ii) What is the direction of the electric field at point P due to charge q2?

______ ° counterclockwise from the +x axis

(iii) Is the magnitude of the electric field at P equal to E1 + E2, where E1 and E2 are the magnitudes of the electric fields produced by q1 and q2?

Yes or No

Explain.

E1 and E2 are scalar quantities.    

Calculations:
(a) What is the net electric field at P?

(b) A small object of charge q0 = +9.00 µC and mass m = 2.0 g is placed at P. When it is released, what is its acceleration?

Explanation / Answer

(i) Electric field is independent of the test charge hence there will be an electric field.

(ii) The force is correctly shown in figure b as the vector E2 so the direction is 310 counterclockwise from x-axis.

(iii) The magnitude is not just a scalar sum of the two vectors (modulus) rather their vector sum. So the answer is no.

(a)

The angle between the two vector (as shown in the figure b) is: = 1800 - 310 - 310 = 1180

So the net electric field will be E = {E12 + E22 + 2*E1*E2*cos}0.5

where E1 = E2 = 9*109*2.8*10-6/0.7*0.7 = 5.143*104 N/C

Hence E = 5.143*104*1.414*(1 + cos118)0.5 = 5.29*104 N/C

Also the direction will be 900 degree counterclockwise to the x-axis (as shown in figure b)

(b)

acceleration here will be given by: a = q0*E/m = 9*10-6*5.29*104/2*10-3 = 238.05 m/s2

and the direction will be same as that of the electric field i,e  900 degree counterclockwise to the x-axis.

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