The force exerted by one plate for two parallel plates arrangement is F = QE/2.
<h3>
Force one plate exerts on the other</h3>
The force exerted by one plate for two parallel plates arrangement is given as follows;
F = QE/2
where;
- Q is the charge on one plates
- E is the electric field due to the charge
- F is force exerted by one plate
Thus, the force exerted by one plate for two parallel plates arrangement is F = QE/2.
Learn more about force between parallel plates here: brainly.com/question/13590045
Answer:
Explanation:
Given the simultaneous equation,
3C+4D=5 .............. 1
2C+5D=2 ............... 2
Solving for the value of C and D using substitution method.
From equation 1;
3C = 5-4D
Divide both sides by 3
3C/3 = (5-4D)/3
C = (5-4D)/3 .... 3
From equation 2:
2C+5D=2
5D = 2-2C
Divide both sides by 5;
5D/5 = 2-2C/5
D = (2-2C)/5 ..... 4
Substitute equation 4 into 3;
C = 5-4{(2-2C)/5}/3
C = [5 - (8-8C/5)]/3
C = [25-(8-8C)/5]/3
C = (17+8C)/15
15C = 17+8C
15C-8C = 17
7C = 17
C = 17/7
Substitute C = 17/7 into equation 4 to get the value of D
D = (2-2(17/7))/5
D = (2-34/7)/5
D = 14-34/35
D = -20/35
D = -4/7
<em>Hence the value of C = 17/7, D = -4/7</em>
Answer:
Explanation:
Difference in the measurement by shuttle crew Δ d = 238857 - 226316 miles
= 12541 miles
Actual measurement = 238857 miles
percentage error
= (12541 / 238857 ) x 100
= 5.25 %
Answer:
This link was diagram
Explanation:
https://doubtnut.app.link/FnsNC80Dccb
Answer:
The work done is 120000 J.
Explanation:
Given that,
Weight = 600 N
Distance = 200 m
We need to calculate the work done
Using formula of work done

Where, F = force
d = distance
Put the value into the formula


Hence, The work done is 120000 J.