ELectric Charges

For this Question I tried Kq1q2/r^2 = -K(q1 + 1)(q2 +2) / r^2

the condition Came -q1q2 = q1 + q2 but non of the options satisfy

iTeachChem Helper•8/10/24, 6:06 PM

@Gyro Gearloose

iTeachChem Helper•8/10/24, 6:06 PM

Note for OP

+solved @user1 @user2...

+solved @user1 @user2...

to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.

GamertugOP•8/10/24, 6:06 PM

force same but opp dircn so F = -F

GGamertug For this Question I tried Kq1q2/r^2 = -K(q1 + 1)(q2 +2) / r^2 the condition Cam...

Ishizora•8/10/24, 7:03 PM

-q1q2 = ( q1 +1 ) ( q2 + 1 )
q1q2 + q1 + q2 + 1 = -q1q2
2q1q2 + q1 + q2 + 1 = 0
satisfies option c

GGamertug force same but opp dircn so F = -F

Ishizora•8/10/24, 7:05 PM

nope you can't just apply that directly
carefully looking at the question it says initially forces where attractive which mean they had opposite charges
Later the forces b/w them became repulsive which means that they now had same sign
applying what i've mentioned above you'll get your answer as option c

Ishizora•8/10/24, 7:15 PM

btw it can be done by simple elimation as well
at first charges should of opposite sign to be attractive
it is given that mag remains same so just check options by multipying them before adding 1 and after adding one if they remain same then that will be your answer which in this case is c