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jok3333 [9.3K]
2 years ago
11

A point charge is placed 3m from a 4uc charge what is the strength of the electric field on the point charge at this distance ro

und to the nearest thousand
Physics
1 answer:
Vlada [557]2 years ago
7 0

The strength of the electric field on the point charge at this distance will be 4000 V/m.

<h3>What is the strength of the electric field?</h3>

The strength of the electric field is the ratio of electric force per unit charge.

The given data in the problem is;

Qis the unit charge = 4.0 × 10⁻⁶ C

E is the strength of the electric field

R is the distance from point charge = 3 m

The strength of the electric field is;

\rm E = \frac{KQ}{R^2} \\\\ \rm E = \frac{9 \times 10^9 \times 4 \times 10^{-6} \ C}{3^2} \\\\ E= 4000 V/m

Hence, the strength of the electric field on the point charge at this distance will be 4000 V/m.

To learn more about the strength of the electric field refer to the link;

brainly.com/question/15170044

#SPJ1

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In a parallel circuit the current A stays in one path B splits and goes through two components C makes one circle
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Answer:

B splits and goes through two components

Explanation:

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- A parallel circuit is a circuit consisting of separate branches, so that each branch has a potential difference equal to the emf of the battery. As a result, in such a circuit the current in the circuit splits and goes through the different branches/components.

So, the correct answer is

B splits and goes through two components

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A typical wall outlet voltage is 120 volts in the United States. Personal MP3 players require much smaller voltages, typically 2
butalik [34]

Answer:        

Number of turns in secondary will be 7

Explanation:

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We have to find the number of turns in secondary

We know that \frac{N_p}{N_s}=\frac{V_p}{V_s}

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Help with #3 please
Marizza181 [45]

Answer:

The answer to your question is va = 8 cm/s, vb = 12.5 cm/s, a = 9 cm/s²

Explanation:

Data

Ta = 0.125 s

Tb = 0.08 s

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distance = 1 cm

Process

1.- Calculate va

va = 1/0.125 = 8 cm/s

vb = 1/0.08 = 12.5 cm/s

2.- Calculate Δv

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3.- Calculate acceleration

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