Answer: electric potential difference
Explanation:
Answer:
The highest electric field is experienced by a 2 C charge acted on by a 6 N electric force. Its magnitude is 3 N.
Explanation:
The formula for electric field is given as:
E = F/q
where,
E = Electric field
F = Electric Force
q = Charge Experiencing Force
Now, we apply this formula to all the cases given in question.
A) <u>A 2C charge acted on by a 4 N electric force</u>
F = 4 N
q = 2 C
Therefore,
E = 4 N/2 C = 2 N/C
B) <u>A 3 C charge acted on by a 5 N electric force</u>
F = 5 N
q = 3 C
Therefore,
E = 5 N/3 C = 1.67 N/C
C) <u>A 4 C charge acted on by a 6 N electric force</u>
F = 6 N
q = 4 C
Therefore,
E = 6 N/4 C = 1.5 N/C
D) <u>A 2 C charge acted on by a 6 N electric force</u>
F = 6 N
q = 2 C
Therefore,
E = 6 N/2 C = 3 N/C
E) <u>A 3 C charge acted on by a 3 N electric force</u>
F = 3 N
q = 3 C
Therefore,
E = 3 N/3 C = 1 N/C
F) <u>A 4 C charge acted on by a 2 N electric force</u>
F = 2 N
q = 4 C
Therefore,
E = 2 N/4 C = 0.5 N/C
The highest field is 3 N, which is found in part D.
<u>A 2 C charge acted on by a 6 N electric force</u>
Answer:
26945.6 ft⋅lbf
Explanation:
Volume of Right Circular Cone = pi*(radius^2)*(height/3)
Pi*(4)*(5/3) = 20.94 ft^3
Density = Mass / Volume
Mass = Density*Volume
Mass = (40)*(20.94)
Mass = 837.6 lb
Work = Force*Height
Force = Mass*Acceleration
Acceleration will be gravitational acceleration
Work = (837.6)*(32.17)*(1)
Work = 26945.6 ft⋅lbf