It states that the electric flux across any closed surface is directly proportional to the net electric charge enclosed by the surface.

$Q = \dfrac {ER^2}k$

Where,

$E$ = electric force = 4286 N/C

$k$ = Coulomb constant = $8.99 \times 10^9 \rm\ N m ^2 /C ^2$

$Q\\
$ = charges = ?

$r$ = distance of separation = 2.5 m

Put the values in the formula,

$Q = \dfrac {4286\times 2.5 ^2}{8.99 \times 10^9 }\\\\
Q = 3\rm \ \mu C$

Therefore, the magnitude of the source charge is 3 μC.

Learn more about Gauss's law:

brainly.com/question/1249602

8 0

2 years ago

In a darkened room, a burning candle is placed 1.84 m from white wall. A lens is placed between candle and wall at a location th

KATRIN_1 [288]

Answer:

82.4 cm

Explanation:

The object and screen are kept fixed ie the distance between them is fixed and by displacing lens between them images are formed on the screen . In the first case let u be the object distance and v be the image distance

then ,

u + v = 184 cm

In the second case of image formation , v becomes u and u becomes v only then image formation in the second case is possible.

The difference between two object distance ie( v - u ) is the distance by which lens is moved so

v - u = 82.4 cm

7 0

3 years ago

Formula of relative displacement

Zolol [24]

what is the question?

8 0

3 years ago