How do you find the direction and magnitude from a nucleus?

[itsmarasilly]

Homework Statement

The nucleus of an atom has a charge of 65 protons.

(a) What are the direction and magnitude of the electric field at 4.9 x 10^-10 m from the nucleus?

(b) What are the direction and magnitude of the force exerted on an electron located at this distance?

Homework Equations

E= KQ/d^2
K= 9 x 10^9

The Attempt at a Solution

(a) (9 x 10^9)(65)/(4.9 x 10^-10)^2
= 2.44e30 N

Please help clarify if (a) is correct, and what exactly is (b) asking, if any different from (a)?

 

[LowlyPion]

I think you need to understand what the charge of 65 protons is.
http://en.wikipedia.org/wiki/Proton

It is not 65 coulombs that are the units of the equation you used.

Having found the field, then b) is asking you to find the force.

 

[nlightner]

I would approach this problem by first understanding the concepts of electric field and force. The electric field is a measure of the force per unit charge at a given point in space. It is a vector quantity, meaning it has both direction and magnitude. The direction of the electric field at a point is defined as the direction of the force that a positive test charge would experience if placed at that point. The magnitude of the electric field is given by the equation E= KQ/d^2, where K is the Coulomb's constant, Q is the charge of the nucleus, and d is the distance from the nucleus.

(a) To find the direction and magnitude of the electric field at a distance of 4.9 x 10^-10 m from the nucleus, we can use the equation E= KQ/d^2. Plugging in the given values, we get E = (9 x 10^9)(65)/(4.9 x 10^-10)^2 = 2.44 x 10^30 N/C. This means that at a distance of 4.9 x 10^-10 m from the nucleus, a positive test charge would experience a force of 2.44 x 10^30 N in the direction away from the nucleus.

(b) The question is asking for the direction and magnitude of the force exerted on an electron located at the same distance from the nucleus. Since an electron has a negative charge, it would experience a force in the opposite direction of the electric field. The magnitude of this force can be calculated using the equation F= qE, where q is the charge of the electron and E is the electric field. Plugging in the values, we get F = (1.6 x 10^-19)(2.44 x 10^30) = 3.9 x 10^11 N. This means that at a distance of 4.9 x 10^-10 m from the nucleus, an electron would experience a force of 3.9 x 10^11 N towards the nucleus.

In summary, to find the direction and magnitude of the electric field and force at a given distance from a nucleus, we can use the equations E= KQ/d^2 and F= qE respectively. It is important to note that the direction of the electric field and force are opposite for positive and negative charges.