67 solve pls 67 find the electric field and electrotatic potential profile due to the charge dis

First, let's determine the electric field due to the charge distribution. We can use Gauss's Law

67 solve pls 67 find the electric field and electrotatic...

6.7 Find the electric field and electrostatic potential profile due to the charge distribution shown in Fig. P6.7. Make labeled sketches of x and the potential x. In your potential plot, take the reference potential to be = 0 at x = 0. 20 -2W W W Charge distribution FIGURE P6.7 130 MICROELECTRONIC DEVICES AND CIRCUITS Vo Impulse (Number per cm2) . N - X

Transcript

00:01 In this problem, we have been given that there are two charges.

00:03 Q and q they are kept at the positions indicated, and we need to figure out the total electric field's magnitude at this point b, provided that a is 60 centimeter, b is 80 centimeters, and charge capital q is minus 6 nanoculum, which is minus 6 into 10 raise to minus 9 column, and small q charge, that's 3 into 10 raise to minus 9 column.

00:28 So here we will use the expression of electric field that's kq by r square to get the electric field because of one point charge.

00:36 Since q is negative, we know the direction of electric field.

00:39 It's from positive to the negative.

00:41 So it will be in this direction.

00:43 And electric field because of this small q that will be in this direction along the line joining because it's positive charge and electric field will be away from it.

00:53 Now here we observe.

00:54 Let's take the electric field because of small charges e and electric field because of the big q charge as so e here will be 9 into 10 raise to 9 that's the value of k we multiply with the magnitude of charge that's 6 into 10 raise to minus 9 and we divide it by the square of the distance here we observe that the angle is 90 degree and using pythagoras rule the distance r can be computed by taking the squares of a and b and taking the root of that so we get that as 100 cm, which is 1 meters.

01:30 So when we square one, it's going to be just the same thing.

01:33 So the electric field comes out to be 54 newton's per column.

01:37 Also, let's figure out the angle here because we will require them to get the complete magnitude.

01:43 So angle can be determined by using simple trigonometry, and that angle will be tan inverse of the perpendicular upon the adjacent.

01:51 So when we take the perpendicular and divide it by the adjacent, we're going to get to.

01:56 0 .75 and we take the tan inverse of that to get the angle here as 37 degree approximately.

02:06 So here the magnitude of this electric field we have figured out and let's convert that in terms of vector.

02:12 So it will have two components, one along the negative x direction and along the positive y direction will be one component.

02:20 So e vector will be minus e cos 37 degree along i cap plus e sine 37 degree along positive wide direction.

02:32 So here when we multiply 54 with the value of cost 37, we're going to get minus 43 .2 icap plus here we multiply the magnitude of e with sign 37, and that's 32 .4j cap...

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