Review What is the strength of the electric field at the...

Review What is the strength of the electric field at the position indicated by the dot in (Figure 1)? E = 9700 N/C Previous Answers Correct Figure 1 of 1 3.0 nC 5.0 cm Part B What is the direction of the electric field at the position indicated by the dot in (Figure 1)? Specify the direction as an angle measured clockwise from the positive x axis. 10 cm -6.0 nC ? = 336.46

Transcript

00:01 Hi, here in this given problem, first of all this is a positive charge having a value of 3 .0 nanoculum and exactly below it there is a negative charge having a value of 6 .m minus 6 .0 nanoculum.

00:23 The gap between them it is marked as it is given as 10 cm and the dot at which we have to find net electric field.

00:38 We assume this upper charge to be at a point a, the lower charge to be at a point b, and the observation point we assume it to be at a point c, whose distance from a that is given as 5 .0 centimeter.

00:57 So at this point c there will be two electric fields.

01:02 One because of a going away another because of this minus 6 .0 nanopulam that will be towards b e b it is marked as now suppose this angle here suppose this is theta and that we will assume to be alpha in the first part of the problem we have to find magnitude of net electric field at point c so first of all this ac distance that is simply 5 .0 centimeter or we can say this is 5 .0 multiplied by 10 to per minus 2 meter and the distance of this point from the lower charge means minus 6 .0 nanoculum this is bc and that will be given by pythagoras theorem means this is square of 10 plus square of 5 means this is 125 centimeter or square root of 125 multiplied by 10 to the power minus 2 meter.

02:28 So first of all, ea using its formula k q by r square.

02:37 This is 9 multiplied by 10 to the power 9 for charge q.

02:41 At point a this is 3 multiplied by 10 to the bar minus 9 for the conversion of nano divided by r square which is 5 multiplied by 10 to the power minus 2 square so this ea is calculated to be equal to 1 .08 multiplied by 10 raised to power 4 newton per coulame now e b that will be given by k q by square of b c means again 9 multiplied by 10 to par 9 multiplied by magnitude of lower charge which is 6 nanoculum or 6 multiplied by 10 to power minus 9 coulam divided by square of bc means 125 multiplied by 10 to bar minus 4 it will become which is calculated to be equal to 0 .432 multiplied by 10 to the power for newton percula.

03:48 Okay, now this angle theta, in this right angle triangle, it will be given by in triangle bac, 10 theta is 10 by 5 means this is 2.

04:07 So theta will be given by 10 inverse 2 which is calculated to be equal to 63 so angle alpha that will be given by 180 degree minus 63 .43 degree and it is calculated to be equal to 116 .6 degrees...

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