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Dafna11 [192]
3 years ago
12

If the charge of a particle doubles, what happens to the force acting on it?

Physics
1 answer:
bazaltina [42]3 years ago
8 0

Explanation:

(1) The force that exists between charged particles is electrostatic force. It is given by :

F=k\dfrac{q_1q_2}{r^2}

Where

q₁ and q₂ are charges

r is distance between charges

If the charge of a particle doubles, the electric force doubles. So, the correct option is (a) "It doubles".

(2) A charge exerts a negative force on another charge. Negative force denotes the force is attractive. It means that the charges are of opposite sign. So, the correct option is (c) "the charges are of opposite signs".

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Two identical charges q placed 2.0 mapart exert forces of magnitude 4.0 N on each other What is the value of the charge q? a) q
katen-ka-za [31]

Answer:

c) 4.2*10^{-5}C

Explanation:

Coulomb's law says that the force exerted between two charges is inversely proportional to the square of distance between them, and is given by the expression:

F=\frac{kq_{1}q_{2}}{r^{2}}

where k is a proportionality constant with the value k=9*10^{9}\frac{Nm^{2}}{C^{2}}

In this case q_{1}=q_{2}=q, so we have:

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

Solving the equation for q, we have:

kq^{2}=Fr^{2}

q^{2}=\frac{Fr^{2}}{k}

q=\sqrt{\frac{Fr^{2}}{k}}

Replacing the given values:

q=\sqrt{\frac{4.0N*(2.0m)^{2}}{9*10^{-9}\frac{Nm^{2}}{C^{2}}}}

q=4.2*10^{-5}C

3 0
3 years ago
An automobile starter motor has an equivalent resistance of 0.0510 Ω and is supplied by a 12.0 V battery with a 0.0090 Ω interna
Alisiya [41]

Answer:

a) 200A

b) 10.2V

c) 2.04kW

d)

I=80A

V=4.08V

P=0.326kW

Explanation:

Here we have a circuit of one power source and two resistors in series, the first question is asking for the current, so according to Ohm's Law:

I=\frac{V}{R}

Where R is the equivalent resistance of the resistors in series

R=0.0510+0.0090=0.0600[ohm]

I=\frac{12.0}{0.0600}=200A

To calculate the voltage dropped by the motor we have to apply the voltage divider rule:

V_m=V*\frac{R_m}{R_m+R_s}\\V_m=12.0*\frac{0.0510}{0.0600}\\V_m=10.2V

The power dissipated supplied to the motor is given by:

P=I^2*R_m\\P=(200)^2*0.0510=2.04kW

now solving adding a 0.0900 ohm resistor:

I=\frac{12.0}{0.15}=80A

V_m=12.0*\frac{0.0510}{0.15}\\V_m=4.08V

P=I^2*R_m\\P=(200)^2*0.0510=0.326kW

5 0
3 years ago
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