3see E=E(x)i+E(y)j+E(z)k
-∂V/∂r=E(x)i+E(y)j+E(z)k
where r is a radial vector of the point. is this wat u are asking.
or r=xi+yj+zk
the sum is
there is a point charge of +2.4 micro C located at the origion.....
find the work done in displacing a 2nd pnt charge of q=4.30micro C from x=0.150 y=0 to x=0.250 to y=0.250
my doubt is that we can first easily sove it by te electic field methd
ex=distance form the x axis and in y axis distance form ht ey axis
thereofe f= F=q/4\pi\varepsilon (r1-r2)/(r1-r2)^{}3
where the top r1 and r2 signify vecotrs and ds and we know dx
but how wil we use when we are solving by potential it is a scalar quantity now we know the distanc efrom pythogoras theorem
but suppose in the question only x and y cordinates are given then
i am assuming the the potential will be the sm of the potnial in the x cordinate +y cordinate
pls see this and is my working for the work by the force method rite????
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4 Answers
23yeah ..work done is independent of path taken .....in a conservative field ...
1ya thats y doubt but i slved
if the objects cordintaes are x1 and y1
and the potnetial at pnt x2 nd y2 is aske
then potential is
calculated by the distance formula from cordinate geomety
i am talking here abt the r
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