Three-point charges are placed on the x-y plane: a +35.0 nC...

Three-point charges are placed on the x-y plane: a +35.0 nC charge at the origin, a -35.0 nC charge on the x axis at 10.0 cm, and a +120 nC charge at the point (10.0 cm, 8.00 cm). 1. Find the y component of the total electric force on the +120 nC charge due to the other two. (Express your answer to two significant figures.) 2. Find the y component of the electric field at the location of the +120 nC charge due to the presence of the other two charges? (Express your answer to two significant figures.)

Transcript

00:01 Hello students, in the question we are given three charges, we have to find total electric force on charge q3 due to charge q1 and q2.

00:12 So here due to charge q1 we would have a repulsive force because both are positive charges and due to a charge q2 we would have an attractive force that is towards charge q2 because they are opposite charges.

00:29 Here let us find out this angle theta, here we have tan theta is equal to this distance that is 8 centimeter divided by this distance that is 10 centimeter so this is 4 by 5 from here we get theta is equal to tan inverse of 4 by 5 this comes out as 38 .659 degree.

00:57 Also here we have r1 is equal to under root 8 square plus 10 square by pythagoras theorem because this angle it is 90 degree we get r1 equal to 12 .806 centimeter.

01:17 Now let us find out electric force force on charge q3 due to charge q1 that is f13, here f13 it is equal to 1 upon 4 pi epsilon naught q1 q3 magnitude of charges divided by r1 square and we have r1 cap unit vector we have here f13 it is equal to 9 into 10 to the power 9 newton meter square per coulomb square value of this constant into charges 35 .0 into 10 to the power minus 9 coulomb into 120 into 10 to the power minus 9 coulomb divided by r1 square that is 12 .806 into 10 to the power minus 2 meter square into here we have cos theta i cap plus sin theta j cap unit vector r1 cap here we get f13 is equal to 2 .304 into 10 to the power minus 3 newton into cos of 38 .659 degree i cap plus sin of 38 .659 degree.

02:56 Now let us find out electric force on q3 due to charge q2 here we have f23 it is equal to 1 upon 4 pi epsilon naught q2 q3 divided by r2 square minus r2 cap vector so here we get f23 is equal to 9 into 10 to the power 9 newton meter square per coulomb square into 35 .0 into 10 to the power minus 9 coulomb into 120 into 10 to the power minus 9 coulomb divided by 8 .0 into 10 to the power minus 2 meter square minus j cap and we have here f23 equal to 5 .906 into 10 to the power minus 3 newton minus j cap...

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