Consider the charge configuration as shown in the figure. Calculate the electric field at point A. If an electron is placed at points A, what is the acceleration experienced by this electron? (mass of the electron `= 9.1 xx 10^(-31)` kg and charge of electron `= -1.6 xx 10^(-19)" C"`)

Explanation

By using superposition principle, the net electric field at point A is
`vec(E)_(A)=(1)/(4piepsilon_o) (q_1)/( r_(1A)^2) hat(r)_(1A) + (1)/(4pi epsilon_(o) ) (q_2)/( r_(2A)^(2) ) hat(r)_(2A)`
where `r_(1A) and r_(2A)` are the distances of point A from the two charges respectively.
`vec(E)_(A) = (9 xx 10^(9) xx 1 xx 10^(-6) )/( (2 xx 10^(-3) )^2) (hat(j) ) + (9 xx 10^9 xx 1 xx 10^(-6) )/( (2 xx 10^(-3) )^2) (hat(i))`
`2. 25 xx 10^(9) hat(j) + 2.25 xx 10^(9) hat(i) = 2.25 xx 10^(9) (hat(i) - hat(j) )`
The magnitude of electric field
`|vec(E)_(A)| sqrt( (2.25 xx 10^9)^2 + (2.25 xx 10^9)^2) `
`= 2.25 xx sqrt(2) xx 10^9 "NC"^(-1)`
The direction of `vec(E)_(A)` is given by
`( vec(E)_A) /( | vec(E)_(A) |) = (2.25 xx 10^9 (hat(i ) + hat(j) ))/( 2.25 xx sqrt(2) xxx 10^9) = ( (hat(i) + hat(j) )/( sqrt2))`, which is the unit vector along OA as shown in the figure.

The acceleration experienced by an electron placed at point A is
`vec(E)_(A) = (vec(F))/( m) = (qvec(E)_A)/(m)`
`= ((-1.6 xx 10^(-19) ) xx (2.25 xx 10^9) (hat(i) + hat(j) ) )/( 9.1 xx 10^(-31) )`
`= -3.95 xxx 10^(20) (hat(i) + hat(j) )N`
The electron is accelerated in a direction exactly opposite to `vec(E)_A`.