Two charges are placed on the x axis 700 c at the origin and...

Two charges are placed on the x-axis, +7.00 µC at the origin and -10.0 µC at x = 10.0 cm. Define the positive direction as the +x-direction. Find the electric field $\vec{E}$ on the x-axis at x = 6.00 cm. $\vec{E}$ = N/C At what point(s) $x_0$ on the x-axis is the electric field zero? If there are multiple points, enter the x-coordinates as a comma-separated list. Round values to three significant figures. $x_0$ = m

Transcript

00:01 Hello and welcome to this video solution of numerate.

00:04 So here it's given that two charges are placed on the x -axis.

00:08 Positive 8 mu -culomb at the origin.

00:11 So here we have origin, let's say, x equal to 0.

00:15 And here we have q1.

00:17 Q1 is 8 mu -culomb, right? and there is this minus 10 mu -culomb at x -equal -to -10.

00:26 So here we have x -equal to 10 centimeters, which we have q2, right? q2 is negative 10 mu -culum right here now you have to take the positive x direction to be positive and you have to find out the electric field at x equal to 6 cm so here we have x equal to 6 right right so this is a positive charge plus q1 and this is negative right so obviously what you can do here is the electric field due to q1 is directed to the right because it is moving away from this positive charge and q2 is towards it again the electric field so e1 and e2 will be directed towards q to charge right so the net e will be equal to e1 plus e2 right and this is equal to k q1 over r1 square plus k q2 over r2 square and you have to only take the magnitude of q1 and q2 so k i can take common and plug in the value 9 times of 10 to the power of 9 q1 is 8 mu -culum so 8 times of 10 to the part of minus 6 over r1 is 6 cm right which is 0 .06 squared plus this is 10 times of 10 to the power of minus 6 over now you have got the distance between this 6 and 10 as 4 centimeters right so 0 .04 squared so let's see how much we're getting so the net electric field will be 7 .63 times of 10 to the power of 7 newton per column is a electric field at x equal to 6 right next you have to find out the point where the electric field is 0 so let's this part b so let's look into this situation now q1 is having a lesser magnitude than q2 8...

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