Answer:
The necessary separation between the two parallel plates is 0.104 mm
Explanation:
Given;
length of each side of the square plate, L = 6.5 cm = 0.065 m
charge on each plate, Q = 12.5 nC
potential difference across the plates, V = 34.8 V
Potential difference across parallel plates is given as;

Where;
d is the separation or distance between the two parallel plates;

Therefore, the necessary separation between the two parallel plates is 0.104 mm
Answer:
the electric field strength on the second one is 2.67 N/C.
Explanation:
the electric fiel on the first one is:
E1 = k×q/(r^2)
r^2 = k×q/(E1)
= (9×10^9)×(q)/(24.0)
= 375000000q
then the electric field on the second one is:
E2 = k×q/(R^2)
we know that R = 3r
R^2 = 9×r^2
E2 = k×q/(9×r^2)
= k×q/(9×375000000q)
= k/(9×375000000)
= (9×10^9)/(9×375000000)
= 2.67 N/C
Therefore, the electric field strength on the second one is 2.67 N/C.
The correct answer is C. Final Velocity
Hope this helped!
Answer:
80m
Explanation:
u=20,R=?,sin theta=1,g=10
R=u²sin2theta/g
R=20²x2/10
R=400x2=800/10
R=80m