± Suspending Charged Particles Using Electric Fields Part A What must the charge

Question

± Suspending Charged Particles Using Electric Fields

Part A

What must the charge (sign and magnitude) of a particle of mass 1.45 g be for it to remain stationary when placed in a downward-directed electric field of magnitude 650 N/C ?

Use 9.81 m/s2 for the magnitude of the acceleration due to gravity.

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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 1.67×1027 kg for the mass of a proton, 1.60×1019 C for the magnitude of the charge on an electron, and 9.81 m/s2 for the magnitude of the acceleration due to gravity.

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Explanation / Answer

(1) Mass of particle = m = 1.45 g=1.45*10^-3 kg

For partical to remain stationary , the force due to field should be upwards.

Electric field = E = 650 N/C downward-directed

As electric field is downward-directed , the force on charge will be upwrds if charge is negative.

force due to field = weight

qE = mg

q = mg / E =1.45*10^-3*9.8 /650

q =2.18*10^-5 C

The charge is negative of magnitude 2.218*10^-5C
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(2 )
For proton,

E =mg/q=1.67*10^-27*9.81/1.6*10^-19=1.0239... N/C

The magnitude of electric field is 1.02391875*10^-7 N/C

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