Tipler6 21.P.032. A point charge of-2 C is located at the origin. A second point

Question

Tipler6 21.P.032. A point charge of-2 C is located at the origin. A second point charge of 14-C is at x-1 m, y-05 m. Find the x and y coordinates of the position at which an electron would be in equilbim 12x= m Submit You currently have 3 submissions for this question. Only 10 submission are allowed. You can make 7 more submissions for this question. Your submissions: 0.607 x Computed value: 0.607 Submitted: Thursday, March 8 at 9:13 PM Feedback: 2) y Submit You currently have 0 submissions for this question. Only 10 submission are allowed. You can make 10 more submissions for this question

Explanation / Answer

The electron must be somewhere on the line joing the 2 charges, and at a positon where net force on it to be zero, it must be nearer to -ve than +ve charge; i.e. in the 3rd quadrant. S, let positon be x,y.

Then, repulsive force= attractive force

Or, k*2*e/r1^2= k*14*e/r2^2 ; where r1= (x^2+y^2)^1/2 and r2= ((x-1)^2+(y-0.5)^2)^1/2 { only magnitudes of the charges are taken }

So, 2/r1^2= 14/r2^2

Or, 14*r1^2= 2*r2^2

So, 7r1^2= r2^2

So, 7*x^2+7*y^2 = (x-1)^2+(y-0.5)^2

Or, 6x^2+6y^2-1.25+2x+y = 0

Now, the euqtion of the line passing through both given points is 2y= x. So, putting x=2y in 6x^2+6y^2-1.25+2x+y = 0;

24y^2+6y^2-1.25+4y+y = 0

30y^2+5y-1.25= 0

So, y= -0.304 and x= -0.607

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