Question

A A square of4cm side having +2coulomb
charge on a
on a each corner. Find thepotential energy
of this system.
​

Answer 1

Answer:

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Explanation:

Solution :

Method 1 (using direct formula)

U=U12+U13+U14+U23+U24+U34

=Kq2a+Kq2a2–√+Kq2a+Kq2a+Kq2a2–√+Kq2a

=[4Kq2a+2Kq2a2–√]=2Kq2a[2+12–√]=a22πε0a[2+12–√]

Method 2 [using U=(U1+U2+...)]

U1 is the total potential energy of charge at corner 1 due to all other charges , U2 is the total potential energy of charge at corner 2 due to all other chargers , U3 is the total potential energy of charge at corner 3 due to all other charges , U4 is the total potential energy of charge at corner 4 due to all other charges.

Since due to symmetry,

U1=U2=U3=mU4

Unet=U1+U2+U3+U42

12.4.[Kq2a+Kq2a+Kq22a−−√]=2Kq2a[2+12–√].