A proton is placed in electric field of strength 6.68×10 power 9 in/cm .Calculate its acceleration
Answers
electric field strength,E = 6.68 × 10^9 N/C, right ?
charge on electron, q = 1.6 × 10^-19 C
mass of electron , m = 9.1 × 10^-31 Kg
at equilibrium,
electrostatic force = Newtonian force
or, qE = ma [ from Newton's 2nd law ]
or, a = qE/m
= (1.6 × 10^-19 × 6.68 × 10^9)/(9.1 × 10^-31)
= 1.174 × 10^(-19 + 9 + 31) m/s²
= 1.174 × 10²¹ m/s²
hence, acceleration of electron is 1.174 × 10²¹ m/s²
[note : if mentioned answer in your book didn't match with given above answer. then you should check unit of electric field strength. maybe your mentioned unit is different. well, it's good news for you. just use above concept in your data and then definitely you can get absolute answer ]
Answer:
Electric fields is the force experienced by a unit positive charge.
Explanation:
Here electric field strength that is E is given to be 6.68/10^9 in/cm
Now we have to calculate acceleration. We know the formula of force. So we will compare the equations of force and arrive at the acceleration value.
E*q=ma
Here, m is the mass of the proton
q is the charge on the proton.
a= E*q/m
a= (6.68*10^9 * 1.6*10^+19 Coulombs)/ 1.6*10^ -27 kg.
I am leaving this to you to evaluate. In these type of questions first find the relation between the asked quantities and known quantities by making different equations. Then solve for that unknown quantity. And take a note for the units as well. Convert it according to your need.