R2 = 0.8 m q1 q2 P x r1 = 1.2 m Fig. Q1 Figure Q1 shows two...

r2 = 0.8 m q1 q2 P x r1 = 1.2 m Fig. Q1 Figure Q1 shows two point charges located on the x-axis, in a medium whose relative permittivity, ?r = 2. Charge q1 = +0.6 ?C, charge q2 = -0.5 ?C. What is the magnitude of the electric field E at the point P due to the two charges? What is the direction of the E field (i.e., in the +x or -x direction)? Note that 1 ?C = 10-6 C.

Transcript

00:01 Hello, in the question we have given a figure which contains two point charges located on the x -axis.

00:08 So, this is basically the x -axis in a medium whose relative permittivity, so e r, epsilon r is given which is 2.

00:19 So, the charge q1, so this is the charge q1 which is of 0 .6 microcoulomb and charge q2 is minus of 0 .0 microcoulomb and the separation, the distance of separation is given.

00:34 So, we have to find out the electric field at point p.

00:37 So, due to this two charges, so let's say e1 is the electric field due to this q1.

00:43 So, it will be in this direction that is e1 and now due to this q2 charge there will be an electric field but it will be in this direction because it is minus charge.

00:55 So, the electric field is towards the negative charge.

00:59 So, this is we are calling it as e2.

01:02 So, what will be e1? so, e1 will be equal to 1 divided by 4 pi epsilon because we have now a medium into q1 divided by r1 square and what is the direction? it is in the positive x cap.

01:21 So, now this 1 upon 4 pi, this epsilon we can write it as epsilon not epsilon r q1 by r1 square in the x cap direction.

01:34 Now, let us plug the value.

01:36 Now, we know that 1 upon 4 pi epsilon not this quantity is 9 into 10 raised to 9.

01:41 Then q1 is given as 0 .6 into 10 raised to minus 6 coulombs divided by this epsilon r is 2 into r1.

02:00 See what is r1? 1 .2 square.

02:05 So, if we calculate this, we will get this e1 as equal to so it will be 5400 divided by so 1 .2 square into 2 is 2 .88 in the x cap direction.

02:21 So, if we calculate this, we will get this e1 as equal to 1875 in the x cap direction.

02:32 So, this is my e1.

02:34 Similarly, e2 will be equal to e2 is due to the point charge q2.

02:38 So, it will be 1 upon 4 pi epsilon into q2 divided by r2 square in the minus x cap direction because see where we have shown this electric field, it is going in the negative x direction...

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