Physics, asked by mustafeez51, 1 year ago











: Calculate the electric field intensity E
which would be just sufficient to balance the weight of
electron. If this electric field is produced by a second
lectron located below the first one what would be the
distance between them ? [Given: e = 1.6 x 10-19 C.
m= 9.1 x 10-31kg and g = 9.8 m/s?]
ution : As force on a charge e in an electric field E.

Answers

Answered by akanksha135
19

Eq = mg

Where E is electric field intensity, q is the charge, m is mass and g is acceleration due to gravity.

E = mg/q

=9.1×10^-31 × 9.8 /1.6×10^-19

=46.9×10^-12

Answered by jhangir789
0

The electric field E is 46.9×10^-12.

​What is an electric field simple definition?

  • Electric field, an electric property associated with each point in space when charge is present in any form.
  • The magnitude and direction of the electric field are expressed by the value of E, called electric field strength or electric field intensity or simply the electric field.

What is electric field SI unit?

  • The SI unit of the electric field is volts per meter (V/m).
  • This unit is equivalent to Newton's per coulomb.
  • These are derived units where Newton is a unit of force and Coulomb is the unit of charge.

According to the question:

Eq = mg

Where E is electric field intensity, q is the charge, m is mass and g is acceleration due to gravity.

E = mg/q

=9.1×10^-31 × 9.8 /1.6×10^-19

=46.9×10^-12

Hence, The electric field E is 46.9×10^-12.

Learn more about electric field here,

https://brainly.in/question/776253?msp_poc_exp=5

#SPJ2

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