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2 years ago

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two charges having the same charge magnitude experiencing an attracting force of 3.60N when the charges are 30cm apart.what is t

he charge magnitude in micro coulombs and the sign of each charge? Q1= q2=

1 answer:

Tomtit [17]2 years ago

3 0

The charges have opposite sign and magnitude $6 \mu C$

Explanation:

The magnitude of the electrostatic force between two electric charges is given by Coulomb's law:

$F=k\frac{q_1 q_2}{r^2}$

where:

$k=8.99\cdot 10^9 Nm^{-2}C^{-2}$ is the Coulomb's constant

$q_1, q_2$ are the two charges

r is the separation between the two charges

In this problem, we have:

F = 3.60 N is the force between the two charges

r = 30 cm = 0.30 m is their separation

The two charges have same magnitude, so

$q_1 = q_2 = q$

So we can rewrite the equation as

$F=\frac{kq^2}{r^2}$

And solving for q:

$q=\sqrt{\frac{Fr^2}{k}}=\sqrt{\frac{(3.60)(0.30)^2}{8.99\cdot 10^9}}=6\cdot 10^{-6} C = 6\mu C$

Moreover, the force between the charges is attractive: we know that charges of same sign repel each other while charges of opposite sign attract each other, therefore the charges in this problem have opposite sign, so

$q_1 = 6 \mu C\\q_2 = -6 \mu C$

Learn more about electric force:

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3 years ago

How much momentum does a 77 kg football player have if he is running with a speed of 4 m/s

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3 years ago

A +71 nC charge is positioned 1.9 m from a +42 nC charge. What is the magnitude of the electric field at the midpoint of these c

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Answer:

The net Electric field at the mid point is 289.19 N/C

Given:

Q = + 71 nC = $71\times 10^{- 9} C$

Q' = + 42 nC = $42\times 10^{- 9} C$

Separation distance, d = 1.9 m

Solution:

To find the magnitude of electric field at the mid point,

Electric field at the mid-point due to charge Q is given by:

$\vec{E} = \frac{Q}{4\pi\epsilon_{o}(\frac{d}{2})^{2}}$

$\vec{E} = \frac{71\times 10^{- 9}}{4\pi\8.85\times 10^{- 12}(\frac{1.9}{2})^{2}}$

$\vec{E} = 708.03 N/C$

Now,

Electric field at the mid-point due to charge Q' is given by:

$\vec{E'} = \frac{Q'}{4\pi\epsilon_{o}(\frac{d}{2})^{2}}$

$\vec{E'} = \frac{42\times 10^{- 9}}{4\pi\8.85\times 10^{- 12}(\frac{1.9}{2})^{2}}$

$\vec{E'} = 418.84 N/C$

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$\vec{E_{net}} = 708.03 - 418.84 = 289.19 N/C$

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3 years ago

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3 years ago

Assuming the diode is ideal but with a forward voltage drop of. 65 volts, what is the current in ma if v1=0v, and r1=490ω?

weeeeeb [17]

The current in the ideal diode with forward biased voltage drop of 65V is 132.6 mA.

To find the answer, we have to know more about the ideal diode.

<h3>What is an ideal diode?</h3>

A type of electronic component known as an ideal diode has two terminals, only permits the flow of current in one direction, and has less zero resistance in one direction and infinite resistance in another.
A semiconductor diode is the kind of diode that is used the most commonly.
It is a PN junction-containing crystalline semiconductor component that is wired to two electrical terminals.

<h3>How to find the current in ideal diode?</h3>

Here we have given with the values,

$V_2=65V\\V_1=0V\\R_1=490Ohm.$

We have the expression for current in mA of the ideal diode with forward biased voltage drop as,

$I=\frac{V_2-V_1}{R_1} =\frac{65}{490} =132.6mA$

Thus, we can conclude that, the current in mA of the ideal diode with forward biased voltage drop of 65 V is 132.6.

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1 year ago