Question

if equal charges are placed on each cube corner. find force on any one charge due to 7 other...​

Answer 1

Answer:

The net force particle a can be given by vector sum of force experienced by this particle due to all the other charges on vertices of the cube. For this we use vector from of coulomb's law F→=kq1q2∣∣r→1−r→2∣∣3(r→1−r→2)

From the figure the different force acting on A are given as F→A1=kq2(−akˆ)a3

F→A2=kq2(−ajˆ−akˆ)(2–√a)3,F→A3=kq2(−aiˆ−ajˆ−akˆ)(3–√a)3,F→A4=kq2(−aiˆ−akˆ)(2–√a)3

F→A5=kq2(−aiˆ)a3,F→A6=kq2(−aiˆ−ajˆ)(2–√a)3,F→A7=kq2(−ajˆ)a3

The net force experienced by A can be given as

F→net=F→A1+F→A2+F→A3+F→A4+F→A5+F→A6+F→A7=−kq2a2[(133–√+12–√+1)(iˆ+jˆ+kˆ)]

Answer 2

Answer:

3a2Kq2

Let, find out force on charge at A.

∴FAB=a2K(q)(q)=a2Kq2

Similarly,

FAC=a2Kq2

Now, Angle b/w FAB&FAC=600

Using parallelogram of vector addition,

Fr1=(FAB)2+(FAC)2+2FABFACcos