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Electric Potential Difference. Let’s revisit another type of Potential energy….
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Let’s revisit another type of Potential energy…. The point to be made is that work must be done by an external force to move an object against nature (WORK IS DONE) - from low potential energy to high potential energy. On the other hand, objects naturally move from high potential energy to low potential energy under the influence of the field force.
In this same manner, to move a charge in an electric field against its natural direction of motion would require work. The exertion of work by an external force would in turn add potential energy to the object. The natural direction of motion of an object is from high energy to low energy; work must be done to move the object againstnature
On the other hand, work would not be required to move an object from a high potential energy location to a low potential energy location.
Now we will consider the motion of the same positive test charge within the electric field created by a negative source charge. The same principle regarding work and potential energy will be used to identify the locations of high and low energy.
The electric potential difference is defined in terms of the work to be done on a charge to move it against an electric field. The electric potential V is a scalar quantity defined as the potential energy per unit charge. V = W qo W = Vqo Units: J/C = Volt (V)
POTENTIAL AND ELECTRIC FIELD The potential difference between two oppositely charged plates is equal to the product of the field intensity and the plate separation. W = q VBA W = F d = q Ed (d is distance) VBA = E d Units: VBA = Volts (V) E = V/m or N/C
CAPACITORS (look on PSE page 212) A capacitor is a device that can store electric charge and consists of two conducting objects placed near one another but not touching. A typical capacitor consists of a pair of parallel plates of area A separated by a distance d. Often the two plates are rolled into the form of a cylinder with paper or other insulator.
The capacitance of a parallel-plate capacitor can be found by using this equation: εo = 8.85x10-12 C2/Nm2 To increase capacitance… -increase area of plates -decrease gap between
If a voltage is applied to a capacitor it becomes charged. The amount of charge acquired by each plate is proportional to the voltage. C is the capacitance in farads(F), Q is the charge in (C) V is the voltage (V) I like to remember as Q=VC
Applications of Capacitors: Keyboards and Defibrillators Charging… clear…
SAMPLE The capacitance of a capacitor is 1.27x10-10 F whose plates are 12 cm x 12 cm separated 1 mm by an air gap.Find the charge on each plate if the capacitor is connected to a 12 V battery. Q = CV = 1.27x10-10 (12) = 1.53x10-9 C
Example: A 127 μF capacitor is connected to a 12 V battery, find the charge on each plate. Q = CV = 127 x 10-6 (12) = 1.52 x 10-3 C practice problems