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Nonamiya [84]
4 years ago
13

Four point charges are individually brought from infinity and placed at the corners of a square. Each charge has the identical v

alue +Q. The length of the diagonal of the square is 2a. What is the electric potential at P, the center of the square?
a. kQ/4a
b. kQ/a
c. zero volts
d. 2kQ/a
e. 4kQ /a
Physics
1 answer:
Brut [27]4 years ago
3 0

To solve this problem we will apply the concept of voltage given by Coulomb's laws. From there we will define the charges and the distance, and we will obtain the total value of the potential difference in the system.

The length of diagonal is given as

l = 2a

The distance of the center of the square from each of the corners is

r = \frac{2a}{2}= a

The potential electric at the center due to each cornet charge is

V_1 = \frac{kQ_1}{r_1}

V_2 = \frac{kQ_2}{r_2}

V_3 = \frac{kQ_3}{r_3}

V_4 = \frac{kQ_4}{r_4}

The total electric potential at the center of the given square is

V = V_1+V_2+V_3+V_4

V = \frac{kQ_1}{r_1}+ \frac{kQ_2}{r_2}+\frac{kQ_3}{r_3}+\frac{kQ_4}{r_4}

Al the charges are equal, and the distance are equal to a, then

V = \frac{kQ}{a}+ \frac{kQ}{a}+\frac{kQ}{a}+\frac{kQ}{a}

V = \frac{4kQ}{a}

Therefore the correct option is E.

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Calculate the work done by a force of 30 N in lifting a load of 2 kg to a height of 10 m (g = 10 m/s2)
Mars2501 [29]

Answer:

<u>300 J</u>

Explanation:

Given :

  1. Applied force = 30 N
  2. Mass = 2 kg
  3. Height = 10 m
  4. g = 10 m/s²

Work done by the force :

  • Work done = Force x Displacement
  • Work done = mg x h
  • Work done = 30 N x 10 m
  • Work done = <u>300 J</u>

<u></u>

<u><em>Note</em></u> :

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2 years ago
What are forces that two objects apply on each other
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HOPE IT HELPED:) HAVE A NICE DAY
3 0
3 years ago
A barometer accidentally contains 6.5 inches of water on top of the mercury column (so there is also water vapor instead of a va
Alisiya [41]

Answer:

(a). 14.4 lbf/in^2.

(b). 27.8 in, AS THE TEMPERATURE INCREASES, THE LENGTH OF MERCURY DECREASES.

Explanation:

So, from the question above we are given the following parameters which are going to help us in solving this particular Question;

=> The "barometer accidentally contains 6.5 inches of water on top of the mercury column (so there is also water vapor instead of a vacuum at the top of the barometer)"

=> "On a day when the temperature is 70oF, the mercury column height is 28.35 inches (corrected for thermal expansion)."

With these knowledge, let us delve right into the solution;

(a). The barometric pressure = water vapor pressure + acceleration due to gravity (ft/s^2) × water density(slug/ft^3) × {ft/12 in}^3 × [ height of mercury column + specific gravity of mercury × height of water column].

The barometric pressure= 0.363 + {(62.146) ÷ (12^3) × 390.6425}. = 14.4 lbf/in^2.

(b). { (13.55 × length of mercury) + 6.5 } × (62.15÷ 12^3) = 14.4 - 0.603.

Length of mercury = 27.8 in.

AS THE TEMPERATURE INCREASES, THE LENGTH OF MERCURY DECREASES.

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3 years ago
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