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

Answer:

a) x = 0.200 m

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

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

$E = \frac{kq}{x}^2$

according to question E = 0

FROM FIGURE

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

$\frac{k21}{x}^2 = \frac{k47}{0.5-x}^2$

solving for x we get

$\frac{0.5}{x} = 1+ \sqrt{\frac{47}{21}}$

x = 0.200 m

b)electric field at half way mean x =0.25

$E =\frac{k*21*10^{-6}}{0.25^2} -\frac{k*47*10^{-6}}{0.25^2}$

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

Answer 2

Answer:

(a) 0.2 m

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

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

$E_{1}=\frac{K q_{1}}{d^{2}}$

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

$E_{1}=\frac{9\times 10^{10}\times 21\times 10^{-6}}{d^{2}}$

$E_{1}=\frac{189000}{d^{2}}$    ..... (1)

Electric field at P due to charge q2

$E_{1}=\frac{K q_{2}}{(r-d)^{2}}$

$E_{1}=\frac{9\times 10^{10}\times 47\times 10^{-6}}{(r-d)^{2}}$

$E_{1}=\frac{432000}{(r-d)^{2}}$    ..... (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.

$E_{1}=\frac{9\times 10^{9}\times 21\times 10^{-6}}{0.25\times 0.25}$

E1 = 3024 x 10^3 N/C

$E_{2}=\frac{9\times 10^{9}\times 47\times 10^{-6}}{0.25\times 0.25}$

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