Respuesta :
Answer: sheet of charge
Explanation:
a )
Since the charge is negative , potential will be negative near it . At a far point potential will be less negative. So potential will virtually increase on going away from the sheet . At infinity it will become almost zero. Electric field will be towards the plate , so potential will decrease towards the plate.
b ) The shape of equi -potential surface will be plane parallel to the sheet of charge because electric field will be perpendicular to the sheet of charge and almost uniform near the sheet of charge. The equi- potential surface is always perpendicular to electric field.
C ) Electric field which is almost uniform near the sheet of charge is equal t the following
E = σ / ε₀ where σ is charge density of surface and ε₀ is permittivity of medium whose value is 8.85 x 10⁻¹²
E = 3 x 10⁻⁹ / 8.85 x 10⁻¹²
= .3389 x 10³
= 338.9 V / m
spacing between 1 V
= 1 / 338.9 m
= 2.95 X 10⁻3 m
= 2.95 mm.
The effect of the potential, the shape of the equipotential surface and the spacing between surfaces is mathematically given as
- the potential will decrease towards the plate.
- an equipotential surface is always perpendicular
- d= 2.95 mm.
What are the effect of the potential, the shape of the equipotential surface, and the spacing between surfaces?
a ) For this case we are dealing with a negative charge, hence we experience a negative potential, it will become almost zero at infinity. therefore, the potential will decrease towards the plate.
b ) The equipotential surface is always perpendicular to the electric field.
c )
Generally, the equation for the electric field is mathematically given as[tex]E = \sigma / e[/tex]
E = 3 x 10⁻⁹ / 8.85 x 10⁻¹²
E= .3389 x 10³ = 338.9 v/m
In conclusion,the distance in spacing between 1 V is
d= 1 / 338.9 m
d= 2.95 X 10⁻3 m
d= 2.95 mm.
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