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3 years ago

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Two objects separated by a distance r are each carrying a charge q The magnitude of the force exerted on the second object by th

e first is F If the first object is removed and replaced with an identical object that carries a charge 4q what is the magnitude of the electric force on the second object

1 answer:

Sindrei [870]3 years ago

4 0

Answer:F=4F

Explanation: Columbs law states that The force between the two point charges is directly proportional to the product of charges and inversely proportional to the square of distance between them

Force between the two charges is given by

F=K*q1*q2/r^2

if one charge become 4 times, new force is,

F=4(K*q1*q2)/r^2

F=4F

Where q1 and q2 are the point charges

r is the distance between the two charges

K is a constant of proportion called electrostatic force

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What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1

notka56 [123]

Complete Question

Part of the question is shown on the first uploaded image

The rest of the question

What is (Fnet3)x, the x-component of the net force exerted by these two charges on a third charge q3 = 55.0 nC placed between q1 and q2 at x3 = -1.220 m ? Your answer may be positive or negative, depending on the direction of the force. Express your answer numerically in newtons to three significant figures.

Answer:

The net force exerted on the third charge is $F_{net}= 3.22*10^{-5} \ J$

Explanation:

From the question we are told that

The third charge is $q_3 = 55 nC = 55 *10^{-9} C$

The position of the third charge is $x = -1.220 \ m$

The first charge is $q_1 = -16 nC = -16 *10^{-9} \ C$

The position of the first charge is $x_1 = -1.650m$

The second charge is $q_2 = 32 nC = 32 *10^{-9} C$

The position of the second charge is $x_2 = 0 \ m$

The distance between the first and the third charge is

$d_{1-3} = -1.650 -(-1.220)$

$d_{1-3} = -0.43 \ m$

The force exerted on the third charge by the first is

$F_{1-3} = \frac{k q_1 q_3}{d_{1-3}^2}$

Where k is the coulomb's constant with a value $9*10^{9} \ kg\cdot m^3\cdot s^{-4}\cdot A^2.$

substituting values

$F_{1-3} = \frac{9*10^{9}* 16 *10^{-9} * (55*10^{-9})}{(-0.43)^2}$

$F_{1-3} = 4.28 *10^{-5} \ N$

The distance between the second and the third charge is

$d_{2-3} = 0- (-1.22)$

$d_{2-3} =1.220 \ m$

The force exerted on the third charge by the first is mathematically evaluated as

$F_{2-3} = \frac{k q_2 q_3}{d_{2-3}^2}$

substituting values

$F_{2-3} = \frac{9*10^{9} * (32*10^{-9}) *(55*10^{-9})}{(1.220)^2}$

$F_{2-3} = 1.06*10^{-5} N$

The net force is

$F_{net} = F_{1-3} -F_{2-3}$

substituting values

$F_{net} = 4.28 *10^{-5} - 1.06*10^{-5}$

$F_{net}= 3.22*10^{-5} \ J$

6 0

2 years ago

In a certain clock, a pendulum of length L1 has a period T1 = 0.95s. The length of the pendulum

gulaghasi [49]

Answer:

Ratio of length will be $\frac{L_2}{L_1}=1.108$

Explanation:

We have given time period of the pendulum when length is $L_1$ is $T_1=0.95sec$

And when length is $L_2$ time period $T_2=1sec$

We know that time period is given by

$T=2\pi \sqrt{\frac{L}{g}}$

So $0.95=2\pi \sqrt{\frac{L_1}{g}}$ ----eqn 1

And $1=2\pi \sqrt{\frac{L_2}{g}}$ -------eqn 2

Dividing eqn 2 by eqn 1

$\frac{1}{0.95}=\sqrt{\frac{L_2}{L_1}}$

Squaring both side

$\frac{L_2}{L_1}=1.108$

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3 years ago

A cook removes a one-gallon pot of hot soup from a stove and places it in an ice-water bath to cool. which is the best cooling p

Verizon [17]

I thank it is #2 or #1 hop this helps;)

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3 years ago

Analyzing the Light Bulb: You should have noticed that the light bulb doesn't have a single well-defined "resistance," since the

VLD [36.1K]

Answer:

Resistance increases with increase in temperature which depends on power supplied which also depends on voltage.

Thermal expansion will make resistance larger.

Explanation:

Light bulb is a good example of a filament lamp. If we plot the graph of voltage against current we will notice that resistance is constant at constant temperature.

The filament heats up when an electric current passes through it, and produces light as a result.

The resistance of a lamp increases as the temperature of its filament increases. The current flowing through a filament lamp is not directly proportional to the voltage across it.

tensile stress begins to appear in resistor as the temperature rises. Thus, the resistance value increases as the temperature rises. Resistance value can only decrease as the temperature rises in case of thin film resistor with aluminium substrate.

In case of a filament bulb, the resistance will increase as increase in length of the wire. The thermal expansion in this regard is linear expansivity in which resistance is proportional to length of the wire.

Resistance therefore get larger.

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3 years ago

A baseball player throws a baseball at a speed of 40 meters per second at an angle of 30 degrees. What is the velocity of projec

Vesna [10]

Answer:

The horizontal component of the velocity is the cosine of 30 degrees multiplied by 40m/s. The cosine of 30 degrees is the 0.8660 . To get the speed, multiply by 40m/s. This equals 34.64, which is approximately 35m/s.

Hope it helpss :)

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2 years ago

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