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2 years ago

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Two equal, but oppositely charged particles are attracted to each other electrically. The size of the force of attraction is 48.

1 N when they are separated by 60.9 cm. What is the magnitude of the charges in microCoulombs?

1 answer:

galina1969 [7]2 years ago

6 0

Given:

The force of attraction is F = 48.1 N

The separation between the charges is

$\begin{gathered} r=\text{ 60.9 cm} \\ =\text{ 60.9}\times10^{-2}\text{ m} \end{gathered}$

Also, the magnitude of charge q1 = q2 = q.

To find the magnitude of charge.

Explanation:

The magnitude of charge can be calculated by the formula

$\begin{gathered} F=\frac{k(2q)}{r^2} \\ q=\frac{Fr^2}{2k} \end{gathered}$

Here, k is the Coulomb's constant whose value is

$k\text{ = 9}\times10^9\text{ N m}^2\text{ /C}^2$

On substituting the values, the magnitude of charge will be

$\begin{gathered} q=\frac{48.1\times(60.9^\times10^{-2})^2}{2\times9\times10^{^9}} \\ =9.91\times10^{-10}\text{ C} \\ =9.91\times10^{-4}\text{ }\mu C \end{gathered}$

Thus, the magnitude of each charge is 9.91 x 10^(-4) micro Coulombs.

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2 years ago

Two charges, one of 2.50μC and the other of -3.50μC, are placed on the x-axis, one at the origin and the other at x = 0.600 m

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3 years ago

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4 years ago

Read 2 more answers

umka2103 [35]

Answer:

$x=2.4365\ m$

and

$x=-1.4365\ m$

Explanation:

Given:

first charge, $q_1=5\times 10^{-3}\ C$

second charge, $q_2=3\times 10^{-3}\ C$

position of first charge, $x_1=-2\ m$

position of second charge, $x_2=-1\ m$

Now since there are only 2 charges and of the same sign so they repel each other. This repulsion will be zero at some point on the line joining the charges.

<u>Now, according to the condition, electric field will be zero where the effects of field due to both the charges is equal.</u>

$E_1=E_2$

since first charge is greater than the second charge so we may get a point to the right of the second charge and the distance between the two charges is 1 meter.

$\frac{1}{4\pi.\epsilon_0} \frac{q_1}{(r+1)^2} =\frac{1}{4\pi.\epsilon_0} \frac{q_2}{(r)^2}$

$\frac{5\times 10^{-3}}{(r+1)^2} = \frac{3\times 10^{-3}}{(r)^2}$

$3(r^2+1+2r)=5r^2$

$2r^2-6r-3=0$

$r=3.4365 \&\ r=-0.4365$

Since we have assumed that the we may get a point to the right of second charge so we calculate with respect to the origin.

$x=-1+3.4365=2.4365\ m$

and

$x=-1-0.4365=-1.4365\ m$

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3 years ago

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