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Hello everyone.
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In this particular question initially we are given with two charges q1 and q2 and distance between these two charges is given as 4 .5 cm now there is one more charge which is equally placed symmetrically placed from these two charges at a distance of 3 cm from each charge.
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So this distance is also 3 cm and this distance is also 3 centimeter and we are given that this particular charge whose magnitude is q is moving in the upward direction with some acceleration.
00:32
So this would be possible, we are given that q charge is negative.
00:36
So this would be possible if this particular charge is positive and this charge is negative.
00:40
Okay, so let us assume that the force by which this particular charge is attracting is f1 and the biforce by which this particular charge is reeling this charge is f2.
00:51
Okay, now we are given with values of some unknown constants as the value of charge is given as minus 2 .35, microculum, milliculum i can say, milliculum and this can be written as minus 2 .35 into 10 x to power minus 3 column okay and we are given that the mass of this particular charge is 5 gram and the acceleration of charge in the upward direction is given as 354 meter per second square and in this question we have to find out the value of q1 and q2 basically but for this for solving this question we have to made a assumption that magnitude of q1 and magnitude of q2 is equal to q.
01:32
This is must because we are having two variables and we will be having only one equation so finding both of these charges and different different magnitude will not be possible so we will be considering that magnitude of both the charges are same but there i can say nature is different one is positive and second one is negative now first let us assume that with this particular line which is symmetric to both of these charges let's say this angle is theta so from this particular figure i can can write that sign of theta will be perpendicular divided by hypotenous so this can be 4 .5 divided by 2 which is 2 .25 cm divided by hypoteness which is 3 cm so from here the angle is coming out as 48 .6 degrees okay this is the angle so let us first calculate the value of f net so f net will be f1 vector plus f2 vector okay so first let us calculate uh not just calculate f1 vector we will i will be directly writing the f -night vector.
02:32
So this will be given by f -1 vector which is the force between q1 and q.
02:37
And since i considered q1 as q only, so i can write this as k -qq divided by r -square and the vector is along this direction.
02:46
So i can dissolve into two components, one cost -teta and one sine -teta...