Solutions to Questions on Electric Fields - Handout 1 ... - Cyberphysics

<strong>Solutions</strong> <strong>to</strong> <strong>Questions</strong> on Electric Fields - Handout 1
Q1. Charge Q acts as a point charge <strong>to</strong> create an electric field. Its strength, measured a distance
of 30 cm away, is 40 N/C. Determine the magnitude of the electric field strength at the following
distances from chare Q
a) 60 cm 10 N/C
b) 15 cm 160 N/C
c) 90 cm 4.4 N/C
d) 3 cm 4000 N/C
e) 45 cm 18 N/C
Hints on how <strong>to</strong> 'reason out' the answer:
The electric field strength is inversely related <strong>to</strong> the square of the distance. So by whatever fac<strong>to</strong>r d
changes by, the E value is altered in the inverse direction by the square of that fac<strong>to</strong>r.
The specifics are as follows:
a) d increases by a fac<strong>to</strong>r of 2; divide the original E by 4.
b) d decreases by a fac<strong>to</strong>r of 2; multiply the original E by 4.
c) d increases by a fac<strong>to</strong>r of 3; divide the original E by 9.
d) d decreases by a fac<strong>to</strong>r of 10; multiply the original E by 100.
e) d increases by a fac<strong>to</strong>r of 1.5; divide the original E by (1.5) 2 .
Q2. Charge Q acts as a point charge <strong>to</strong> create an electric field. Its strength, measured a distance
of 30 cm away, is 40 N/C. What would be the electric field strength:
a) 30 cm away from a source with charge 2Q? 80 N/C
b) 30 cm away from a source with charge 3Q? 120 N/C
c) 60 cm away from a source with charge 2Q? 20 N/C
d) 15 cm away from a source with charge 2Q? 320 N/C
e) 150 cm away from a source with charge 0.5Q? 0.80 N/C
Hints on how <strong>to</strong> 'reason out' the answer:
The E value is directly related <strong>to</strong> the source charge and inversely related <strong>to</strong> the square of the
distance. Alter E by the same fac<strong>to</strong>r that the charge changes by; and alter E by the inverse square of
the fac<strong>to</strong>r that d is changed by.
The specifics are as follows
a) Twice the source charge will double the E value.
b) Three times the source charge will triple the E value.
c) Two changes are required: double E since the source charge doubled and divide by 4
since the distance increased by a fac<strong>to</strong>r of 2.
d) Two changes are required: double E since the source charge doubled and multiply by 4
since the distance decreased by a fac<strong>to</strong>r of 2.
e) Two changes are required: divide E by 2 since the source charge halved and divide by
25 since the distance increased by a fac<strong>to</strong>r of 5.
These questions were adapted from questions on the site www.physicsclassroom.com (2011)

Q3. Copy and complete the following table.
<strong>Solutions</strong> <strong>to</strong> <strong>Questions</strong> on Electric Fields - Handout 1
General points:
1) the E value will always be equal <strong>to</strong> the F / q ratio.
2) Any alteration in q (without altering Q and d) will not effect the E value.
3) If q is altered by some fac<strong>to</strong>r, F is altered by that same fac<strong>to</strong>r; but if Q and d are not
changed, the E will not be changed.
4) In the last two rows, the values in red can be any number provided that the F/q ratio is equal
<strong>to</strong> the E value.
Specific points are as follows:
a) Find E by calculating F / q (both of which are given).
b) Find F by multiplying E by q (both of which are given).
c) Find E by calculating F/q (both of which are given).
d) Find F by multiplying E by q (both of which are given).
e) First find E, reasoning that since Q and d are the same in this row as the previous row, the E
value must also be the same. Then find q by multiplying the given value of F by your calculated
value for E.
f) Find F by multiplying E by q (both of which are given).
g) First find E, reasoning that since Q and d are the same in this row as the previous row, the E
value must also be the same. Then find F by multiplying the calculated value of E by the given
value of q.
h) First find E, reasoning that since Q and d are the same in this row as the previous row, the E
value must also be the same. Then find q by multiplying the given value of F by your calculated
value for E.
i) Any value of q and F can be selected provided that the F/q ratio is equal <strong>to</strong> the given value of
E.
j) First find E, reasoning that since Q and d are the same in this row as the previous row, the E
value must also be the same. Then any value of q and F can be selected provided that the F/q
ratio is equal <strong>to</strong> the determined value of E.
These questions were adapted from questions on the site www.physicsclassroom.com (2011)

<strong>Solutions</strong> <strong>to</strong> <strong>Questions</strong> on Electric Fields - Handout 1
Q4. In the table above, identify at least two rows that illustrate that the strength of the electric
field vec<strong>to</strong>r is ...
a) directly related <strong>to</strong> the quantity of charge on the source charge (Q).
b) inversely related <strong>to</strong> the square of the separation distance (d).
c) independent of the quantity of charge on the test charge (q).
a) Rows a and c or rows b and d. To illustrate that E is directly related <strong>to</strong> Q, you must find a set
of rows in which Q is altered by some fac<strong>to</strong>r while q and d are constant.
b) Rows d and f or rows c and h. To illustrate that E is inversely related <strong>to</strong> d 2 , you must find a
set of rows in which d is altered by some fac<strong>to</strong>r while q and Q are kept constant.
c) Rows a and b or rows d and e or rows f and g. To illustrate that E is independent of q you
must find a set of rows in which q is altered but Q and d are kept constant.
Q5. Use unit analysis <strong>to</strong> identify whether kg ms -2 C -1 is an acceptable unit for electric field
strength.
Answer: Yes it is.
A kg is a unit of mass and a m/s 2 is a unit of acceleration. So a kg m/s 2 is a unit of force; in fact, it is
equivalent <strong>to</strong> a new<strong>to</strong>n. Replacing the kg m/s 2 with N converts this set of units <strong>to</strong> N/C which is the
standard metric unit of electric field as E=F/Q.
Q6. Balloon A is negatively charged. It is observed that balloon B exerts a repulsive force upon
balloon A. Would the electric field vec<strong>to</strong>r created by balloon B be directed <strong>to</strong>wards B or away from B?
Explain your reasoning.
Answer: Towards B
If balloon B repels balloon A then balloon B must be negatively charged. The electric field vec<strong>to</strong>rs
are always directed <strong>to</strong>wards negatively charged objects. As such, the E vec<strong>to</strong>rs must be <strong>to</strong>wards
balloon B.
These questions were adapted from questions on the site www.physicsclassroom.com (2011)