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Masja [62]
3 years ago
11

The electric field, generated by a point charge, has strength of 180,000 N/C at a point 1.50 cm from the point charge. What is t

he charge, if the field points towards the charge
Physics
1 answer:
cestrela7 [59]3 years ago
7 0

Given Information:  

Electric field = E =  180,000 N/C

Distance = r = 1.50 cm = 0.015 m

Required Information:  

Charge = q = ?

Answer:

charge = q = 4.5x10⁻⁹ C

Explanation:

The electric field is given by

E = kq/r²

Where k is the coulomb constant k = 9x10⁹ Nm²/C²

q = Er²/k

q = 180,000*(0.015)²/9x10⁹

q = 4.5x10⁻⁹ C

Therefore, the charge is 4.5 nC

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3 0
3 years ago
If the charge on the negative plate of the capacitor is 121 nano-Coulomb, how many excess electrons are on that plate? Write you
Julli [10]

Answer:

n = 756.25 giga electrons

Explanation:

It is given that,

If the charge on the negative plate of the capacitor, Q=121\ nC=121\times 10^{-9}\ C

Let n is the number of excess electrons are on that plate. Using the quantization of charges, the total charge on the negative plate is given by :

Q=ne

e is the charge on electron

n=\dfrac{Q}{e}

n=\dfrac{121\times 10^{-9}}{1.6\times 10^{-19}}

n=7.5625\times 10^{11}

or

n = 756.25 giga electrons

So, there are 756.25 giga electrons are on the plate. Hence, this is the required solution.

6 0
3 years ago
A circular loop with radius r is rotating with constant angular velocity ω in a uniform electric field with magnitude E. The axi
inn [45]

Answer:

\Phi_{E} = E\pi r^2 \omega t

Explanation:

The electric flux is defined as the multiple of electric field and the area that the electric field passes through, such that

\Phi_{E} = \vec{E}\vec{A}

When calculating the electric flux, the angle between the directions of electric field and the area becomes important, especially if the angle is changing with time.

The above formula can be rewritten as follows

\Phi_{E} = EA\cos(\theta)

where θ is the angle between the electric field and the area of the loop. Note that, the direction of the area of the loop is perpendicular to the plane of the loop.

If the loop is rotating with constant angular velocity ω, then the angle can be written as follows

\theta = \omega t

At t = 0, cos(0) = 1 and the electric flux through the loop is at its maximum value.

Therefore the electric flux can be written as a function of time

\Phi_{E} = E\pi r^2 \omega t

3 0
3 years ago
A man rides up in an elevator at 12 m. He gains 6500 J of gravitational potential energy. what is the man's mass?
iren2701 [21]

Answer:

We know that potential energy of a body;

= mass(m)× gravitational acceleration(g) × height(h)

Lets find out the mass of the body

P.E. = mgh

=> 6500J = mass × 9.8m/s^2 × 12m

=>6500J = mass × ( 9.8 × 12 ) × ( m/s^2 × m)

=> 6500 Nm = m × 117.6 × m^2 / s^2

=> 6500/117.6 Ns^2/m = mass [°.° Ns^2/m = kg]

=> 55.272 Kg = mass

Therefore the mass of the body = 55.272 kg ~ <em>6</em><em>0</em><em> </em><em>k</em><em>g</em><em> </em>(Ans)

Hope it helps you

6 0
3 years ago
A 20×10⁹charge is moved between two points A andB that are 30mm apart and have an electric potential difference of 600v between
Alona [7]

Answer:

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Explanation:

7 0
3 years ago
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