Question

The electric field at the point x=5.00cm and y=0 points in the positive x direction with a magnitude of 10.0 N/C . At the point x=10.0cm and y=0 the electric field points in the positive x direction with a magnitude of 17.0 N/C . Assume this electric field is produced by a single point charge. Part AFind the charge's location.Part BFind the magnitude of the charge.

Answer 1

(A). The location of the charge is 26.45 cm.

(B). The magnitude of the charge is 51.1 pC.

Step-by-step explanation:

Given that,

Distance in x axis = 5.00 cm

Electric field = 10.0 N/C

Distance in x axis = 10.0 cm

Electric field = 17.0 N/C

Since, q is the same charge, two formulas can be set equal using the two different electric fields.

(a). We need to calculate the location of the charge

Using formula of force

`F = qE`....(I)

Using formula of electric force

`F =(kq^2)/(d^2)`....(II)

From equation (I) and (II)

`qE=(kq^2)/(d^2)`

`E=(kq)/(d^2)`

`q=(E(x-r)^2)/(k)`...(III)

For both points,

`(E(x-r)^2)/(k)=(E(x-r)^2)/(k)`

Put the value into the formula

`(10.0*(x-5.00)^2)/(k)=(17.0*(x-10.0)^2)/(k)`

`10.0*(x-5.0)^2=17.0*(x-10.0)^2`

Take the square root of both sides

`3.162(x-5.0)=4.123(x-10.0)`

`3.162x-3.162*5.0=4.123x-4.123*10.0`

`3.162x-4.123x=-4.123*10.0+3.162*5.0`

`0.961x=25.42`

`x=(25.42)/(0.961)`

`x=26.45\ cm`

(B). We need to calculate the charge

Using equation (III)

`q=(E(x-r)^2)/(k)`

Put the value into the formula

`q=(10.0(26.45*10^(-2)-5.00*10^(-2))^2)/(9*10^(9))`

`q=5.11*10^(-11)\ C`

`q=51.1\ pC`

Hence, (A). The location of the charge is 26.45 cm.

(B). The magnitude of the charge is 51.1 pC.