Question

Two charges, -24 mu C, and +5.8 mu C are fixed in place and separated by 3.6 m. (a) At what spot along a line through the charges is the net electric field zero? Given the distance of the spot to the positive charge in meters (m). (b) what would be the force on a charge of +14 mu C placed on that spot

Answer 1

electric field due to a negative charge is directed towards a charge while field due to a positive charge is directed away from the charge.

so on the line connecting a negative and a positive charge:

at any point lying in between the charges, field direction due to both the charges will be same

so if we have to find out a point where total field is 0, then we have to look at points that do not lie

in between the two charges.

now, field due to a charge is inversely proportional to the square of the distance and directly propertional to the magnitude of charge.

hence to get zero electric field, the point should be closer to the lesser mangiutde charge.

considering all the discussion above, the point where electric field is zero will be closer to 5.8 uC.

let distance of the point from 5.8 uC is d meters.

then distance of the point from the negative charge=3.6+d meters

equating magnitude of electric field by both the charges:

k*24/(3.6+d)^2=k*5.8/d^2

where k=coloumb's constant=9*10^9

then 24*d^2=5.8*(3.6+d)^2

==>24*d^2=5.8*(3.6^2+d^2+7.2*d)

==>d=3.481 meters

hence at a point 3.481 meters from 5.8 uC and 3.6+3.481=7.081 meters from -24 uC, total field will be zero.

part b:

force on a charge in electric field =charge*electric field

as at this point, electric field is zero, then force on the charge will be zero.