Answer:
a) x = 0.200 m
b)E = 3.84*10^{-4} N/C
Explanation:
[tex]q_1 = 21.0\mu C[/tex]
[tex]q_1 = 47.0\mu C[/tex]
DISTANCE BETWEEN BOTH POINT CHARGE = 0.5 m
by relation for electric field we have following relation
[tex]E = \frac{kq}{x}^2[/tex]
according to question E = 0
FROM FIGURE
x is the distance from left point charge where electric field is zero
[tex]\frac{k21}{x}^2 = \frac{k47}{0.5-x}^2[/tex]
solving for x we get
[tex]\frac{0.5}{x} = 1+ \sqrt{\frac{47}{21}}[/tex]
x = 0.200 m
b)electric field at half way mean x =0.25
[tex]E =\frac{k*21*10^{-6}}{0.25^2} -\frac{k*47*10^{-6}}{0.25^2}[/tex]
E = 3.84*10^{-4} N/C
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
[tex]E_{1}=\frac{K q_{1}}{d^{2}}[/tex]
Where, K is Coulomb constant = 9 x 10^9 Nm^2/C^2
[tex]E_{1}=\frac{9\times 10^{10}\times 21\times 10^{-6}}{d^{2}}[/tex]
[tex]E_{1}=\frac{189000}{d^{2}}[/tex] ..... (1)
Electric field at P due to charge q2
[tex]E_{1}=\frac{K q_{2}}{(r-d)^{2}}[/tex]
[tex]E_{1}=\frac{9\times 10^{10}\times 47\times 10^{-6}}{(r-d)^{2}}[/tex]
[tex]E_{1}=\frac{432000}{(r-d)^{2}}[/tex] ..... (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.
[tex]E_{1}=\frac{9\times 10^{9}\times 21\times 10^{-6}}{0.25\times 0.25}[/tex]
E1 = 3024 x 10^3 N/C
[tex]E_{2}=\frac{9\times 10^{9}\times 47\times 10^{-6}}{0.25\times 0.25}[/tex]
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
