How Do You Calculate the Force on a Charge in an Electric Field?

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To calculate the force on a charge in an electric field, use the formula F = qE, where F is the force, q is the charge, and E is the electric field strength. For a charge of +7.50 µC in an electric field with components Ex = 6.40 x 10^3 N/C and Ey = 8.80 x 10^3 N/C, the force can be determined by calculating the x and y components separately and then using vector addition. The magnitude of the force is found using the Pythagorean theorem, while the angle with the +x axis can be calculated using trigonometric functions. The discussion highlights confusion around applying these equations correctly. Understanding the relationship between electric field components and force is crucial for solving such problems.
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Homework Statement



A charge of q = +7.50 µC is located in an electric field.

The x and y components of the electric field are Ex = 6.40 103 N/C and Ey = 8.80 103 N/C, respectively.

(a) What is the magnitude of the force on the charge?
in Newtons

(b) Determine the angle that the force makes with the +x axis.
in degrees°


Homework Equations





The Attempt at a Solution



I'm not sure how to use the x and y components to solve this.

not sure why i couldn't just use an equation like this: force per coulomb of charge is F/q0 = N/C
 
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Use F=EQ to find the forces in the x and y components. Then it is just basic vector addition
 
Thread 'Correct statement about size of wire to produce larger extension'

Thread 'Correct statement about size of wire to produce larger extension'

The answer is (B) but I don't really understand why. Based on formula of Young Modulus: $$x=\frac{FL}{AE}$$ The second wire made of the same material so it means they have same Young Modulus. Larger extension means larger value of ##x## so to get larger value of ##x## we can increase ##F## and ##L## and decrease ##A## I am not sure whether there is change in ##F## for first and second wire so I will just assume ##F## does not change. It leaves (B) and (C) as possible options so why is (C)...
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