# Electric Fields and Force

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Assess. Statements 6.2.1-6.2.8 due Monday, 10/20/14
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Like the gravitational field around masses, an electric field
occupies the space surrounding charged objects.
To test for the presence of an electric field:
o Bring a small positive charge (q) into the space
o Release the small positive charge
• If the charge experiences a force, then we know there is an electric field
present
• No force, no electrical field
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Known as a Test Charge
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Attractive forces
The test charge will accelerate towards the negatively
charged object
The test charge will follow the path of the electric field of
the negative object:
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Electric charges exert forces on other electric charges
through the electric fields
Quantified through Coulomb’s Law:
o The electric force between two point charges, Q 1 and Q2,
is inversely proportional to the square of their separation
distance and directly proportional to the product of the
two charges:
𝑘𝑄1 𝑄2
𝐹=
𝑟2
𝑘=

1
4𝜋𝜀0
= 8.99 &times; 10−9 N&middot;m2&middot;C-2
e0electric permittivity of a vacuum
o (= 8.85 x 10-12 C2&middot;N-1&middot;m-2)
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Two charges, 4.00 mC and 6.00 mC, are placed along a
straight line separated by a distance of 2.00 cm. Find the
force exerted on each charge.
𝑞1 = 4.00 &times; 10−6 𝐶
𝑞2 = 6.00 &times; 10−6 𝐶
𝑟 = 0.0200 𝑚
𝑘𝑄1 𝑄2
𝐹=
𝑟2
(8.99 &times; 10−9 )(4.00 &times; 10−6 )(6.00 &times; 10−6 )
𝐹=
0.02002
𝐹 = 540 𝑁, 𝑟𝑒𝑝𝑢𝑙𝑠𝑖𝑣𝑒 𝑓𝑜𝑟𝑐𝑒
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Two equally charged lightweight balls, q, are suspended from
strings that are each 10.0 cm long. They repel each other
and have a separation distance between the charged
particles is 14.00 cm. Assume the mass of each of the
lightweight balls is 0.575 g. What is the charge on each of
the balls?
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Electric field lines are drawn in the same direction as the
force the small postive test charge would experience at that
point.
𝐹
𝐸=
𝑞
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The electric field strength is defined as the force per unit
charge experienced by a small positive test charge, q.
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The electric field strength is FELT by the charge in the field
itself.
The electric field strength is CAUSED by the charge creating
the field
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Electric field strength depends on the charge that is
creating the field, and it depends on how far away from the
charge the field is being measured:
𝑘𝑄1 𝑄2
𝐹=
𝑟2
𝐹
𝐸=
𝑞
𝐴𝑠𝑠𝑢𝑚𝑒 𝑄2 𝑖𝑠 𝑡ℎ𝑒 𝑠𝑎𝑚𝑒 𝑎𝑠 𝑞
𝑘𝑄𝑞
𝐹
𝑘𝑄
2
𝑟
𝐸= =
= 2
𝑞
𝑞
𝑟
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What is the electric field strength 7.50 cm from a particle
with a charge of 5.00 mC?
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What force would a charge of +1.00 nC experience at that
point?
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The electric field between two parallel plates is 100.0 N&middot;C-1.
What acceleration would a charge of 2.00 mC and mass
1.00 x 10-3 kg experience if placed in this field? (ignore the
weight of the charged mass)
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Imagine an electric field generated by a charge Q, and
consider a positive test charge, q, in that field.
What must be done to move q closer to Q?
Electric Potential is the work done per unit charge to bring a
positive test charge from far away to the some point P in an
electric field created by Q
𝑊
𝑉=
𝑞
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The work done in moving the charge to point P from infinity
increases the electrical potential energy of the test charge.
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The work done in moving a test charge of 2.0 mC from very
far away to a point P is 1.50 x 10-8 J. What is the electric
potential at point P?
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The potential at a point P is 12.0 V and a charge of 3.00 C
is placed there. What is the electric potential energy of the
charge?
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What is the electric potential energy if the charge placed at
P is – 2.00 C?
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Electric potential difference: the work done to move a
charge from point A in an electric field to point B in an
electric field (neither is very far away from the source of the
field).
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Mathematically, it’s the difference of potential energy for
the charge at each of the positions (W = DU)
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Potential difference is the total work done to move the
charge (change in energy)
Potential is the work per unit charge.
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What work must be performed in order to move a charge of
5.00 mC from the negative plate to the positive plate if a
potential difference of 250. V is established between the
plates?
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A charge of 5.00 mC and mass 2.00 x 10-8 kg is shot with an
initial velocity of 3.00 x 102 m&middot;s-1 between two parallel
plates kept at a potential of 200.0 V and 300.0 V,
respectively. (the charge starts at the low potential end)
o What will the speed of the charged mass be when it gets to the
other (higher potential) plate?
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The amount of work needed to move a charge equal to one
electron’s charge through a potential difference of exactly 1
volt.
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What is the work needed to move a charge of +2e across a
potential difference of 2.0 V?
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What is the work needed to move a charge of +3e across a
potential difference of 5.0 V?
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What is the equivalent of 1 eV, measured in Joules?
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1 eV = work to move 1e through 1V
What’s the charge of 1e?
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W = qV=(1.61x10-19 C)&middot;(1 V) = 1.61 x 10-19 J
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