14- Three point charges are located are located as shown in...

14- Three point charges are located are located as shown in the figure. Find a. The total electric field at point A (8 pts) $E_1 = \frac{kq}{r^2} = \frac{(9 \times 10^9)(6 \times 10^{-6})}{(0.03)^2}$ $= 6 \times 10^7 \text{ N/C } (\hat{j})$ $E_2 = \frac{(9 \times 10^9)(9 \times 10^{-6})}{(0.05)^2}$ $= 3.2 \times 10^7 \text{ N/C } (\hat{j})$ $E_3 = \frac{(9 \times 10^9)(-8 \times 10^{-6})}{(0.04)^2}$ $= -4.5 \times 10^7 \text{ N/C } (\hat{i})$

Transcript

00:01 In the first part of this problem, we are going to calculate the electric potential for this given system at point p.

00:06 So we call this potential v.

00:10 The potential is defined as v equals to k -1 divided by y plus k -2 divided by x plus k -q -3 divided by x.

00:31 Can be also written as v equals to k u1 divided by y plus k divided by x into q 2 plus q 2 plus q 3 well let's call it equation number 1 now let's define the y first so y can be written as y equals to using this right angle triangle square root of 1 .25 meter whole square minus into 1 .25 meter divided by 2 whole square so this will give us the value for y as y equals to 1 .08 meter now let's write the values into equation number 1.

01:34 So v is equal to, we have the value for k as 8 .99, multiply by eternal as power 9, newton meter square per column square.

01:49 And then we have q1 as 7 .45, multiply by turnerous power minus 6 column.

02:01 Divided by we have the value for y which we have calculated as 1 .08 meter plus then we have here again as 8 .99 multiplied as power 9 newton meter square per column square into now here we have the okay.

02:36 So we have to write the value of x here which is equal to 1 .25 meter divided by 2.

02:54 And then we have to write here the value for q2 which is equal to 7 point, sorry 2 .75 multiply by 10x power minus 6 column plus into, we have to.

03:14 Have the we have to write the value of q3 as a minus 1 .72 multiply by 10x power minus 6 column so yeah now this will give us the value for v as a v equals to 76 .8 calo volt where this calo is equal to so this is the required answer of the first part of this question...

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