Two point charges are placed as follows q 1 = + 0.25 x10 -9 C at (1,0) and q 2 =

Question

Two point charges are placed as follows q1= + 0.25 x10-9 C at (1,0) and q2 = -2.0 x10-9 C at (1,3)

a) What is the electric field (magnitude and direction) at theorigin? Begin by drawing a diagram (required) that showsschematically the placement of the charges and direction of theelectric field vectors at the origin due to each charges Use k= 9.0x 10-9 Nm2/C2 .

b) What is the electrical potential at the origin?

c) If we assume the electrical potential at zero at infinity (thatis very far away). How much work has to be done to bring a chargeof +1C from very far away (infinity) to the origin?

Explanation / Answer

a) I can't draw the diagram here. Try it. :-) Since the given positions are coordinates, the mathematics will beeasy. We work separately for the x- and y- components. Rememberthat this is electric field which is a vector. electric field for point charge is E=kq/r2 working for q1's E-field at the origin (0,0)         Ex =k(q1)/(0-1)2 =kq1/1    this is actually to the negativex-axis if you draw it         Ey =0     since there is no y-component for theE-field, you'll find if you draw it for q2        Ex =kq2/(0-1)2 = kq2/1  this is to the negative x-axis        Ey =kq2/(0-3)2 =kq2/3    this is to the negativey-axis then, as a vector, we add the x- and y-components        Ex =(-)kq1/1 + (-)kq2/1    the negative inside the parentheses emphasizesthe negative direction      Ex = -2kq1/1               Ey = 0 +(-)kq2/3 = -kq2/3 watch out for the negative now, we have the x- and y- components of the net field, you cancompute for the magnitude and direction        magnitude is   E = ((Ex)2 +(Ey)2)       i think you can work out for this... NOTE:we are working with physical units. Based on your problem, I don'tknow what is the unit of every units in the coordinate ofchoice. If it is not given, assume it to be 1 unit is to 1 meter.(Actually, this solution assumes it is.)        to find the direction,the angle can be found bytan-1(Ey/Ex)... (Basedon your drawing) b. electric potential is a scalar, and its formula for pointcharges is V=kq/r working for q1's potential in the origin (0,0). But to do that, weneed to find r first by using the distance formula r2 =x2 + y2        r = {(0-1)2+(0-0)2} = 1 therefore        V = kq1/1 for q2's        r = {(0-1)2+(0-3)2} = 2 therefore        V = kq2/2 Adding as scalars, the total electric potential at origin is        V = kq1/1 +kq2/2        justsubstitute the values c. With this assumption, the potential energy will then be U = qVwhere q is the new charge and V is the potential at origin. simply,       U = q(kq1/1+ kq2/2) I hope this will help. Just substitute values and be consciouswhat is the unit of every units in the coordinate system because itwas not stated in the problem. Just assume 1 m each. If you find errors in the solution, be happy, you arelearning.

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