(11%) Problem 10: A point charge of 5.9 \"C is placed at the origin (x1 = 0) of a coordinate system, and another charge of 1.4 M: is placed placed on the x- axis at x2 = 0.21 m. $3 50% Part (3) Where on the x-axis can a third charge be placed in meters so that the net force on it is zero? x3 = otzol Hints: for a cancel HOtCXt COSO tan() a ( 7 3 | 9 HOME cotanO asin() acos() E M 4 5 | 6 atan() acotan() sinh() / * 1 2 | 3 a cosh() tanh() cotanh() + - 0 | END ODegrees Radians x/o BACKSPACE I DEL CLEAR Submit I I Feedback I Igive up! I deduction. Hints remaining: _ There are three distinct regions for this problem on the xaxis what are they? -VVhich region must the third charge go in order to make sure the force can vanish? the origin? To start with, consider the negative xaxis; the magnitude of the force from the charge at the origin will always be larger than the charge on the other side. Will the force ever vanish on the negative x-axis? Feedback: _ for a This is the distance to the test charge from the charge q. What is the distance from deduction A 50% Part (b) What if both charges are positive; that is, what if the second charge is 1.4 \"C? Grade Summary Deductions Potential 100% Submissions Attempts remaining: (_ per attempt) detailed View 1 2 (11%) Problem 12: The electric field 1/3 of the way from a charge q, to a charge q2 is zero. jackson, wakyah - jackson_wakyah @columbusstate.edu How many times larger is 92 than q1 based on the previous statement? 92/91 = 71 cancel notext COs() tan() 7 HOME cotan() asin() acos E vv 4 5 6 atan acotan() sinh() * 1 2 3 cosh() tanh() cotanh() + 0 END O Degrees O Radians VO BACKSPACE DEL CLEAR Submit Hint Feedback I give up! Hints: 6 for a 0% deduction. Hints remaining: 0 Feedback: 0% deduction per feedback. Use the formula for the electric field due to a point charge in terms of the charge and the distance from it. -The electric field is a vector, so electric fields add vectorially. -The charges have the same sign, so the fields they produce along the line between them are oppositely directed. The vector sum of the two fields at any point on that line is therefore the difference of the two vectors, so they will cancel each other at some point. -Equate the magnitudes of the two fields at the point of zero net field and solve for the requested ratio in terms of a ratio of the distances of the charges from that point. If the distance of the zero point from q1 is 1/f of the total distance between the charges, then the distance of the zero point from 92 is (f - 1)/f of the total distance. The ratio of the distances is r2/r] =f - 1