Question

The electric field in an xy plane produced by a positively charged particle is 5.77(5.0i + 3.6j) N/C at the point (5.0, 6.3) cm and 113i N/C at the point (2.3, 0) cm. What are the (a) x and (b) y coordinates of the particle? (c) What is the charge of the particle?

Answer 1

at the point (2.3,0) cm , field is 113 N/C along +ve x axis.

as charge is positive, it is on x axis and at a coordinate (x,0) such that x<2.3 cm.

let charge be q.

then k*q/(0.01*(2.3-x))^2=113

where k =coloumb's constant

==>q/(2.3-x)^2=1.2556*10^(-12)....(1)

field at (5,6.3) is directed along (5,6.3)-(x,0)=(5-x,6.3)

vector of the electric field is (5,3.6)=0.57143*(8.75,6.3)

as these two vectors are in same direction,

5-x=8.75

==>x=-3.75 cm

so the charge is at -3.75 cm

from equation 1,

q=1.2556*10^(-12)*(2.3-(-3.75))^2=4.5958*10^(-11) C

so answers are:

part a: x coordinate=-3.75 cm

part b: y coordinate=0 cm

part c: ccharge=4.5958*10^(-11) C