Two charges, -20 C and +4.7 C, are fixed in place and separated by 3.0 m. a. At what spot along a line through the charges is th
e net electric field zero? Give the distance of the spot to the positive charge in meters (m). (Hint: The spot does not necessarily lie between the two charges.) b. What would be the force on a charge of +14 C placed at this spot?
1 answer:
The zero net electric field point is at a point that is 0.98 m away from 4.7C charge.If a 14C charge is placed at this point then, force acted on the charge placed at this point is equal to zero.
Explanation:
Let at A both net electric field is zero then
At A ,E1=E2
E1=k*Iq1I / (d+x)^2
E2=k*Iq2I /x^2
Equating both
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Answer:
Explanation:
We are asked to calculate the force you are applying to a car. According to Newton's Second Law of Motion, force is the product of mass and acceleration. Therefore, we can use the following formula to calculate force.
The mass of the car is 2000 kilograms and the acceleration is 0.5 meters per second squared.
- m= 2000 kg
- a= 0.05 m/s²
Substitute the values into the formula.
Multiply.
Convert the units. 1 kilogram meter per second squared is equal to 1 Newton. Our answer of 100 kilogram meters per second square is equal to 100 Newtons.
You apply <u>100 Newtons</u> of force to the car.