Two-charges-q-1-10-C-and-q-2-5-C-are-placed-on-the-axis-at-A-10-0-cm-and-20-0-cm-respectively-Determine-a-position-between-the-two-charges-where-the-electric-field-intensity-is-0-

Question Number 13049 by tawa tawa last updated on 12/May/17

Two charges q_{1} = 10 μ C and q_{2} = 5 μ C are placed on the axis at A (10, 0) cm

and (20, 0) cm respectively . Determine a position between the two charges

where the electric field intensity is 0 .

Answered by ajfour last updated on 12/May/17

s a y i t i s s o a t x .

a = 10 c m = 0.1 m, q_{1} = 10 μ C

b = 20 c m = 0.2 m, q_{2} = 5 μ C

f o r z e r o i n t e n s i t y o f e l e c t r i c f i e l d

a t (x, 0) :

\frac{q_{1}}{4 π ϵ_{0} {(x - a)}^{2}} = \frac{q_{2}}{4 π ϵ_{0} {(b - x)}^{2}}

\Rightarrow \frac{b - x}{x - a} = \sqrt{\frac{q_{2}}{q_{1}}} = \frac{1}{\sqrt{2}}

b \sqrt{2} - x \sqrt{2} = x - a

x = \frac{a + b \sqrt{2}}{1 + \sqrt{2}} = (a + b \sqrt{2}) (\sqrt{2} - 1)

x = (0.1 + 0.2 \sqrt{2}) (\sqrt{2} - 1)

x = \frac{(1 + 2 \sqrt{2})) (\sqrt{2} - 1)}{10} m

x = \frac{3 - \sqrt{2}}{10} m = 0.1586 m

e l e c t r i c f i e l d i n t e n s i t y i s z e r o a t

(15.86, 0) c m .

Commented by tawa tawa last updated on 12/May/17

God bless you sir .