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Hunter-Best [27]
3 years ago
8

Two tiny particles having charges 20.0 μC and 8.00 μC are separated by a distance of 20.0 cm What are the magnitude and directio

n of electric field midway between these two charges? (k = 1/4πε0 = 9.0 × 109 N • m2/C2)

Physics
1 answer:
Alecsey [184]3 years ago
8 0

Answer:

The magnitude and direction of electric field midway between these two charges is 10.8\times10^{5}\ N/C along AB.

Explanation:

Given that,

First charge q_{1}= 20\mu C

second charge q_{2}= 8\mu C

Distance = 20 cm

We need to calculate the electric field

For first charge,

Using formula of electric field

E_{1}= \dfrac{kq_{1}}{r^2}

Put the valueinto the formula

E_{1}=\dfrac{9\times10^{9}\times20\times10^{-6}}{10\times10^{-2}}

E_{1}=18\times10^{5}\ N/C

Direction of electric field along AB

We need to calculate the electric field

For second charge,

Using formula of electric field

E_{2}= \dfrac{kq_{2}}{r^2}

Put the valueinto the formula

E_{2}=\dfrac{9\times10^{9}\times8\times10^{-6}}{10\times10^{-2}}

E_{2}=7.2\times10^{5}\ N/C

Direction of electric field along AO

We need to calculate the net electric field at midpoint

E_{net}=E_{1}-E_{2}

E_{net}=(18-7.2)\times10^{5}\ N/C

E_{net}=10.8\times10^{5}\ N/C

Direction of net electric field along AB

Hence, The magnitude and direction of electric field midway between these two charges is 10.8\times10^{5}\ N/C along AB.

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<h3>Hello there!</h3>

Here, you are looking for the amount of heat put in for water, at a mass of 187 grams, to change by 80 degrees.

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The equation is Q = mc∆T, where Q is the amount of energy put in to raise the temperature by a certain amount, m is the mass, c is the specific heat capacity, and ΔT is the amount of temperature change.

The information given:

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A steady electric current flows through a wire. If 9.0 C of charge passes a particular spot in the wire in a time period of 2.0
defon

1) Current: 4.5 A

2) Time taken: 4.7 s

Explanation:

1)

The electric current intensity is defined as the rate at which charge flows in a conductor; mathematically:

I=\frac{q}{t}

where

I is the current

q is the amount of charge passing a given point in a time t

For the wire in this problem, we have

q = 9.0 C is the amount of charge

t = 2.0 s is the time interval

Solving for I, we find the current:

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2)

To solve this problem, we can use again the same formula

I=\frac{q}{t}

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In this problem, we have:

I = 3.0 A (current)

q = 14.0 C (charge)

Therefore, the time taken for the charge to move past a particular spot in the wire is

t=\frac{q}{I}=\frac{14.0}{3.0}=4.7 s

Learn more about electric current:

brainly.com/question/4438943

brainly.com/question/10597501

brainly.com/question/12246020

#LearnwithBrainly

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