Question

A charge Q is placed at the centr of the square of side 2.10 cm, at...

A charge Q is placed at the centr of the square of side 2.10 cm, at the corners of which four identical charges q = 7.5 C are placed. Find the value of the charge Q so that the whole system is in equilibrium.
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Answer #1

Given is:-

Side of the square a=2.10 cm or   a= 2.10 × 10-2772

Four identical charges with  q = 7、5C

The diagonal distance is  (V2 × (2.10 × 10-2))771

Now,

Let consider the charge on the top left corner, the forces experienced by this charge are as shown below:-

F d F C qLd

Now the value of the forces are

F,- F-k

by plugging all the values we get

2.10 x 10-2)2

which gives us

15

The resultant of these two forces will be in the direction of Fc and will be

Fod 1.15 x105)(1.15 x 1015)2

or

Fed 1.63 1015Л

Now,

(9 109)(7.5)2 (V2(2.10 x 10-2)2   

which gives us

F8.12 x 1014v

now the total force experienced by the top left charge in the direction of Fc is

F_{net} = F_c + F_{bd}

by plugging all the values we get

F,et = 8.12 × 1014 + 1.63 × 101

which gives us

Fret 2.442x 1015N

Therefore the midpoint charge should be of negative polarity so that in can counter this net force, therefore

109 ) ( Q) (7.5) 2.10x10-22 (9 2.442 x 1015

which gives us

Q =-7.98C

This should be the required charge on to the midpoint charge so that the whole system remains in equilibrium

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