Part A What must the charge (sign and magnitude) of a...

Part A What must the charge (sign and magnitude) of a particle of mass 1.46 g be for it to remain stationary when placed in a downward-directed electric field of magnitude 680 N/C? Use 9.80 m/s² for the magnitude of the free-fall acceleration. Part B What is the magnitude of an electric field in which the electric force on a proton is equal in magnitude to its weight? Use $1.67 \times 10^{-27}$ kg for the mass of a proton, $1.60 \times 10^{-19}$ C for the magnitude of the charge on an electron, and 9.80 m/s² for the magnitude of the free-fall acceleration.

Transcript

0:00 Hi there.

00:01 So for this problem for part a, we are told that what we are asked about what must the charge sign in magnitude of a particle of a mass that is given and that is 1 .42 grams.

00:23 B for it to remain stationary when placed in a downward directed electric field of a magnitude that is given, electric field is equal to 600 newtons per column.

00:42 So with that said, we want the force of gravity, that is the force of gravity that we know is minus the mass times acceleration due to gravity.

01:00 To be a subtly cancelled electric force, which we know, the electric force, in this case is just simply the charge of the particle times minus the electric field.

01:16 Now we use positive as up and negative as down and carefully keep track of the signs while requiring that electric field and the charge to be and the acceleration due to gravity to be positive.

01:33 So with that said, if we equal the electric field, to minus the gravitational force, we will obtain that this is minus the charge times the electric field.

01:50 And then this is equal to minus minus the mass times the acceleration due to gravity.

02:03 So we can cancel these signs in here, and then we will obtain that the charge is just simply the mass times the acceleration due to gravity, minus in here, divided by the electric field.

02:16 Now, this sign in here indicates that it must be a negative charge because we know that all of these values are positive.

02:26 So we just simply substitute the values that we are given.

02:29 We are given 1 .42 grams, so we can transform that into kilograms by just divided that value.

02:41 Divided by 1 ,000...

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