Respuesta :
Answer:
Explanation:
capacitance of capacitor
= ε₀ A / d , A is area of plate , d is plate separation
= (8.85 x10⁻¹² x 1.7² x 10⁻⁴) /( .35 x 10⁻³ )
= 73 x 10⁻¹³ F
Charge = 4.5 x 10⁻⁷
voltage between the plates,
Δ V = charge / capacitance
= 4.5 x 10⁻⁷ / 73 x 10⁻¹³
= .0616 x 10⁶
= 616 x 10² V
To make the vottage double , we will reduce capacitance to half .
To reduce capacitance half we will increase plate distance to double.
new plate distance
= 1.7 x 2
= 3.4 cm
(a) The voltage between the plates will be 616×10² V.
(b) The plate width would double this voltage will be 3.4 cm.
What is a parallel plate capacitor?
It is a type capacitor is in which two metal plates arranged in such a way so that they are connected in parallel and have some distance between them.
A dielectric medium is a must in between these plates. help to stop the flow of electric current through it due to its non-conductive nature.
(a) The voltage between the plates will be 616×10² V.
The capicitance of the capicitor is guven as;
[tex]\rm C= \frac{\epsilon_0A}{d} \\\\ \rm C= \frac{8.85 \times 10^{-12}\times1.7 \times 10^{-4}}{.35\times 10^{-3}} \\\\ \rm C= 73 \times 10^{-13}\F[/tex]
Voltage - capacitance relation is given as;
[tex]\rm \triangle v= \frac{Q}{C} \\\\ \rm \triangle v= \frac{4.5 \times10^{-7}}{73\times10^{-13}} \\\\ \rm \triangle v=0.616 \times 10^6 \\\\ \rm \triangle v=616 \times 10^2 V[/tex]
Hence the voltage between the plates will be 616×10² V.
(b) The plate width would double this voltage will be 3.4 cm.
The capacitance is inversely propotional to the voltage. So to double voltage we have to reduce the capacitance to half.
The distance is also inverse of the capacitance. So that to lower capacitance half we have to increase plate distance to double value.
So the new distance between the plate is;
[tex]\rm d= 1.7 \times 2 = 3.4\ cm[/tex]
Hence the plate width would double this voltage will be 3.4 cm.
To learn more about the parallel plate capacitor refer to the link;
https://brainly.com/question/12883102
