Electric field problems
The following electric field problems will help you sharpen your knowledge of electric field. Things that you absolutely need in order to solve the problems on this lesson.
 Know how to use the formula below:
Electric field formula: E = k
q
/
d^{2}
 Know how to draw the electric field vector.
 Know how do solve math problems that have multiplication and division of exponents.
 Use k = 9 × 10^{9} N.m^{2}/C_{2}
 Convert d to meter before doing the math.
 The electric field unit is N/C
Problem #1
A particle with charge q = 5Q is placed at a distance d from a point A.
1. Calculate the electric field produced at the point A in terms of k, Q and d.
2. Calculate the electric field produced at the point A if Q = 10
^{10} and d = 3 cm.
Solution
When solving electric field problems, you need
to find the magnitude and the direction of the electric field. You
cannot just look for one and forget about the other.
To find the magnitude, use the formula shown above.
To find the direction, put a small positive test charge in the field and then find out if it will be attracted or repulsed.
For
this problem it is very easy to find the magnitude of the electric
field. Just substitute 5Q for q. The next two electric field problems
will take things to the next level.
1.
2.
d = 3 × 10
^{2}
d
^{2} = 9 × 10
^{4}
E = 9 × 10
^{9}
5 × 10^{10}
/
9 × 10^{4}
E =
45 × 10^{1}
/
9 × 10^{4}
E =
45 × 10^{1} × 10^{4}
/
9
E = 5 × 10
^{3} N/C
Put the test charge at point A. The direction of the electric field is shown below.
More challenging electric field problems
Caution
Take your time when reading the following electric field problems. Do not rush it!
Problem #2
Two particles with charges q
_{1} = 4Q and q
_{2} = 8Q are placed at the same distance d from a point A. Charges q
_{1} and q
_{2} and point A are all on the same line.
Calculate the electric field produced at the point A in terms of k, Q and d.
Let E
_{1} be the electric field for q
_{1}
Let E
_{2} be the electric field for q
_{2}
Net electric field = E = E
_{1} + E
_{2}
Since the denominator is the same, we can just add 4Q and 8Q.
4Q + 8Q = 12Q
Put the test charge at point A. The direction of the electric field is shown below with the brown arrow.
Notice how the brown vector is the result of adding the green vector and the orange vector.
Notice also how the size of the orange vector is twice that of the green since 8Q is twice 4Q.
Problem #3
Three particles with charges q
_{1} = 2Q, q
_{2} = 2Q, and q
_{3} = 4Q are placed at the same distance d from a test charge placed at the origin. Charges q
_{1} , q
_{2}, and q
_{3} and point A are arranged as you see below.
Calculate the electric field produced at the point A in terms of k, Q and d.
q
_{1} is positive, so the orange test charge is repulsed by q
_{1}. We show this with the first brown arrow
q
_{2} is negative, so the orange test charge is attracted to q
_{2}. We show this with the second brown arrow.
The direction and size of the vectors E
_{1} and E
_{2} is shown with the red arrow and the net electrical charge = k
4Q
/
d^{2}
Let E
_{3} be the electric field for q
_{3} (Shown with green arrow)
We need to find the net electric charge for E
_{3} (Green arrow) and E
_{1} and E
_{3} (Red arrow)
We need to add all the x components and all the y components for the two vectors (green and red).
When
looking at the graph, I see that the y components are shown in gold and
dark blue. Moreover, I immediately see that the y components add up to
zero.
For both the green and the red vector, the x components is the blue line.
Find the x component of the green vector
For the green vector, cos(60°)=
Line in blue
/
line in green
Line in blue = line in green × cos(60°)
Line in blue = line in green × 0.5
Line in blue = k
4Q × 0.5
/
d^{2}
Line in blue = k
2Q
/
d^{2}
To Find the x component of the red vector, we basically repeat what we
did for the green vector to end up with the exact same answer.
For the red vector, line in blue = k
2Q
/
d^{2}
Add the two x components and the magnitude of this field is
4Q
/
d^{2}
and the direction is along the xaxis to the right
Perhaps the last two electric field problems were hard. Send your questions.

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