Question

Point charges of 21.0 μC and 47.0 μC are placed 0.500 m apart. (a) At what point (in m) along the line connecting them is the electric field zero? (b) Whai is the magnitude (in N/C) and direction of the electrci field halfway between them?

Answer 1

(a) 0.2 m

(b) 3.744 x 10^6 N/C Rightwards

Step-by-step explanation:

q1 = 21 micro Coulomb = 21 x 10^-6 C

q2 = 47 micro Coulomb = 47 x 10^-6 c

r = 0.5 m

(a) Let the electric field is zero at a distance d at point P from the 21 micro Coulomb.

Electric field at P due to charge q1

`E_(1)=(K q_(1))/(d^(2))`

Where, K is Coulomb constant = 9 x 10^9 Nm^2/C^2

`E_(1)=(9* 10^(10)* 21* 10^(-6))/(d^(2))`

`E_(1)=(189000)/(d^(2))` ..... (1)

Electric field at P due to charge q2

`E_(1)=(K q_(2))/((r-d)^(2))`

`E_(1)=(9* 10^(10)* 47* 10^(-6))/((r-d)^(2))`

`E_(1)=(432000)/((r-d)^(2))` ..... (2)

Equate the equation (1) and equation (2), we get

1.496 d = r - d

1.496 d = 0.5 - d

2.496 d = 0.5

d = 0.2 m

(b) Let E1 be the electric field due to q1 at mid point P and E2 be the electric field due to q2 at mid point P.

`E_(1)=(9* 10^(9)* 21* 10^(-6))/(0.25* 0.25)`

E1 = 3024 x 10^3 N/C

`E_(2)=(9* 10^(9)* 47* 10^(-6))/(0.25* 0.25)`

E2 = 6768 x 10^3 N/C

The resultant electric field is

E = E2 - E1 = (6768 - 3024) x 10^3 = 3.744 x 10^6 N/C Rightwards

Answer 2

a) x = 0.200 m

b)E = 3.84*10^{-4} N/C

Step-by-step explanation:

`q_1 = 21.0\mu C`

`q_1 = 47.0\mu C`

DISTANCE BETWEEN BOTH POINT CHARGE = 0.5 m

by relation for electric field we have following relation

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

according to question E = 0

FROM FIGURE

x is the distance from left point charge where electric field is zero

`(k21)/(x)^2 = (k47)/(0.5-x)^2`

solving for x we get

`(0.5)/(x) = 1+ \sqrt{(47)/(21)}`

x = 0.200 m

b)electric field at half way mean x =0.25

`E =(k*21*10^(-6))/(0.25^2) -(k*47*10^(-6))/(0.25^2)`

E = 3.84*10^{-4} N/C