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2 answers:
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Answer:
Electric field will be
Explanation:
We have given potential difference between the two plates = 15000 volt
Distance between the plates =
We have to find the electric field strength between the plates
Potential difference between the plates is equal to
So electric field strength will be
So electric field will be
The applied voltage is 1.41* 10⁴ kV.
Calculation of applied voltage-
The capacitance of parallel plates in the absence of a dielectric is:
C = ∈₀A/d
where, "d" = the spacing between the plates = 0.01 * 10⁻³ m
"A" = the area of the plates = 4 m²
"∈₀" = the constant = 8.85419*10⁻¹² C²/Nm²
So, C₀ =
= 3541.67 *10⁻⁹ F
The dielectric constant, κ= 3.4, is defined as:
κ = C/C₀
The effective capacitance (with the dielectric) is denoted by "C", whereas the original capacitance is denoted by "C₀". (without the dielectric).
As a result, the new capacitance is:
C = κC₀
However, capacitance is connected to a voltage by:
C = Q/V
utilizing the increased capacitance and accounting for "V":
κC₀ = Q/V
V = Q/κC₀ = = 1.41* 10⁴ kV
Therefore the applied voltage is 1.41* 10⁴ kV.
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