Physics, asked by pabitradebata, 1 month ago

charges of +2 micro coulomb,+3micro coulomb and -8micro coulomb are placed at the vertices of an equilateral triangle of side 10 CM. calculate the magnitude of force acting on -8micro coulomb charge due to other two charges.​

Answers

Answered by pranalithool93
0

Answer:

Net electric field at A due to charges at B and C is

E

A

=2E

Ac

sin50

o

=2×9×10

9

×

(0.20)

2

3×10

−6

×

2

3

=

3

×6.75×10

5

AM=

(20)

2

−(10)

2

=

400−100

=10

3

cm

Let the charge at M be q. Charge q should be positive so that there can be repulsion between the charges at A and M.

E

AM

=9×10

9

×

(10

3

/100)

2

q

=3q×10

11

For A to be in equilibrium

E

A

=E

AM

3

×6.75×10

5

=3q×10

11

or q=

4

9

3

×10

−6

=

4

9

3

μC

Explanation:

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Answered by meghna4022
0

Explanation:

Particles of charge +65, +48, and −95 μC are placed in a line. The center one is L = 60cm from each of the others.Calculate the net force on the left charge due to the other two.

Physics

A point charge (m = 1.0 g) at the end of an insulating string of length L = 51 cm (Fig. 16-66) is observed to be in equilibrium in a uniform horizontal electric field of E = 9200 N/C, when the pendulum's position is as shown in Fig. 16-66, with the charge

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