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Chapter 4: Q27E (page 715)

(a) What must the charge (sign and magnitude) of a 1.45-g particle be for it to remain stationary when placed in a downward-directed electric field of magnitude 650 N/C? (b) What is the magnitude of an electric field in which the electric force on a proton is equal in magnitude to its weight?

Short Answer

Expert verified

Charge of the particle is a negative charge and its magnitude is q=-2186脳105C

Magnitude of the electric field is 1.023脳10-7C

Step by step solution

01

Step 1:

Here, (E) is the electric field, (F) is the electric force on charge q, and (q) is the magnitude of the point charge.

Therefore, the electric field is

E=Fq

And F=mg

Putting the values in

q=mgE=1.45脳10-39.8650=2.186脳10-5C

As force is upward, so (q) is negative.

Therefore, charge of the particle is a negative charge and its magnitude isq=-2186脳105C.

02

Step 2:

As,

mproton=1.6726脳10-27kgqproton=1.60217脳10-19C

So, the equation

E=FqF=mgE=mgq

Putting values,

=1.6726脳10-279.81.60217脳10-19E=1.023脳10-7C

Therefore, the Magnitude of the electric field is 1.023脳10-7C.

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