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AfilCa [17]
2 years ago
11

A source charge generates an electric field of 4286 N/C at a distance of 2. 5 m. What is the magnitude of the source charge? (Us

e k=) 1. 2 µC 3. 0 µC 1. 2 C 3. 0 C.
Physics
1 answer:
svp [43]2 years ago
8 0

The magnitude of the source charge is 3 μC which generates 4286 N/C of the electric field. Option B is correct.

What does Gauss Law state?

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

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3 years ago
1. You place an object 63 cm in front of a converging lens, with a 40 cm focal length.
NikAS [45]

Answer:

1.

109.6 cm ,  - 1.74 , real

2.

1.5

Explanation:

1.

d₀ = object distance = 63 cm

f = focal length of the lens = 40 cm

d = image distance = ?

using the lens equation

\frac{1}{f} = \frac{1}{d_{o}} + \frac{1}{d}

\frac{1}{40} = \frac{1}{63} + \frac{1}{d}

d = 109.6 cm

magnification is given as

m = \frac{-d}{d_{o}}

m = \frac{-109.6}{63}

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2

d₀ = object distance = a

d = image distance = - (a + 5)

f = focal length of lens = 30 cm

using the lens equation

\frac{1}{f} = \frac{1}{d_{o}} + \frac{1}{d}

\frac{1}{30} = \frac{1}{a} + \frac{1}{- (a + 5)}

a = 10

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