The electric field at the point x-10.0 cm and y = 0 points in the positive x dir

Question

The electric field at the point x-10.0 cm and y = 0 points in the positive x direction with a magnitude of 10.0 N/C. At the point x-15.0 cm and y = 0 the electric field points in the positive x direction with a magnitude of 15.0 N/C. Assume that this electric field is produced by a single, point charge (a) Find the location of the point charge. (x, y)s ( 10.1 Your response differs from the correct answer by more than 10%. Double check your calculations. cm (b) Find the sign and magnitude of its charge 2.36-11X Your response differs from the correct answer by more than 10%. Double check your calculations. C

Explanation / Answer

Given

electric field at x = 10 cm, y= 0 cm is 10 N/C and

electric field at x = 15 cm,y= 0 cm is 15 N/C  

and the field is directed in the positive x direction

to satisfy the above conditions the point of location of the point charge should be beyond (15,0)cm let it is x cm from origin and the charge is -ve

we know for +ve charge the field lines are radially outward and -ve charge the field lines are radially inward

we know tha electric field is E = kq/r^2

a)

E1 = kq/(x-0.1)^2 =10 N/C

E2 = kq/(x-0.15)^2 = 15 N/C

E1/E2 = (x-0.15)^2 /(x-0.1)^2 = 10/15

solving for x , we get x = 12.75 cm , x = 37.25 cm

the point x = 12.75 cm is not possible (it does not satisfy the given condition)

b) E1 = kq/(x-0.1)^2 =10 N/C

10 = 9*10^9*q/(0.3725 -0.1)^2

q = 82.50*10^-12 C

q = 82.50 pC

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