Electric potential

Electric potential

• 1.
  SIR JOHNEL V. ESPONILLA GENERALPHYSICS II
• 2.
   Relate theelectric potential with work, potential energy, and electric field.  Determine the electric potential function at any point due to highly symmetric continuous charge distributions.  Infer the direction and strength of electric field vector, nature of the electric field sources, and electrostatic potential surfaces given the equipotential lines.  Define the work done by the electric force.  Calculate the electric field in the region given a mathematical function describing its potential in a region of space.  Solve problems involving electric potential energy and electric potentials.
• 5.
  DIRECTIONS: Check whatyou have learned about electric fields and charges in the previous lesson. Write CHARGE if the statement is correct; otherwise write FIELD in your paper. 1. The SI unit of electric charge is the coulomb, symbol C. 2. Electrostatic effects occur when electrical charges are separated. 3. All charge separation involves the expenditure of energy. 4. The magnitude of the electrostatic force between charges increases as their separation decreases. CHARG E CHARG E CHARG E CHARG E
• 6.
  POTENTIAL ENERGY ----- GRAVITATIONAL
• 7.
  If we wantto move the test charge to plate B, a force must be applied to push against the force of the
• 8.
  If the forceapplied is conservative, the work done can be expressed as POTENTIAL ENERGY U.
• 9.
  If Wab ispositive, Ua is greater than Ub, ∆U is negative, potential energy decreases. Whether the test charge in the e-field is positive or negative, the potential energy increases if the test charge moves OPPOSITE to the direction of the electric force. The potential energy decreases if the test charge moves in the SAME direction of the electric force.
• 10.
  where r isthe distance of separation, qqo are the values of two charges, electric potential energy is expressed as Nm or Joules (J) For electric potential energy with several test charges, this is expressed as
• 11.
  A point chargeq1 = +2.80𝜇C is at origin. How far should the second point charge of +5.20 𝜇C be placed to have electric potential energy of 0.600 J?
• 12.
  A point chargeq1 = +2.80𝜇C is at origin. How far should the second point charge of +5.20 𝜇C be placed to have electric potential energy of 0.600 J? given unknown sketch Q1 = +2.80 x 10-6 C ; Q2 = +5.20 x 10-6 C ; U = 0.600 J r
• 13.
  A point chargeq1 = +2.80𝜇C is at origin. How far should the second point charge of +5.20 𝜇C be placed to have electric potential energy of 0.600 J? strategy solution Rearrange the formula and solve for r since it is unknown. Then, plug-in for the values. 𝑟 = 1 4𝜋𝜖0 𝑞𝑞0 𝑈 r = 1 4𝜋𝜖0 (2.80 𝑥10−6𝐶)(5.20𝑥10−6𝐶) 0.600 𝐽 r = 9 x 109 𝑁𝑚2 𝐶2 1.456𝑥10−11𝐶2 0.600 𝑁𝑚 r = 0.2184 m
• 14.
  A charge of4.50 x 10-8C is placed in a uniform electric field that is directed vertically upward with a magnitude of 5.00 x 104 N/C. What work is done by the electrical force when the charge moves 0.450 m to the right? 0.800 m downward? 2.60 m at an angle of 45 degrees from the horizontal? given unknown sketch Q1 = +4.50 x 10-8 C ; E = 5.00 x 104 N/C ; r = 0.450 m, r = 0.800 m r = 2.60 m, 450 W
• 15.
  A charge of4.50 x 10-8C is placed in a uniform electric field that is directed vertically upward with a magnitude of 5.00 x 104 N/C. What work is done by the electrical force when the charge moves 0.450 m to the right? 0.800 m downward? 2.60 m at an angle of 45 degrees from the horizontal? strategy solution Use W = Fd cos𝜃, where F = qE. W = Fd = qEd = (4.50 x 10-8 C)(5.00 x 104 N/C)(0.45 m)(cos90) = 0J W = Fd = qEd = (4.50 x 10-8 C)(5.00 x 104 N/C)(0.800 m)(cos0) = -1.8 x 10-3 J W = Fd = qEd = (4.50 x 10-8 C)(5.00 x 104 N/C)(2.60 m)(cos45) = 4.137 x 10-3 J
• 16.
  where U isthe electric potential energy and q0 is the charge, electric potential is expressed as volt (V) Divide both sides of the equation relating work done by the electric force from points a to b by q0 to represent it as work per unit charge.
• 17.
  The difference Va– Vb is known as the potential difference. To find the electric potential V of a single point charge, express as 1. The potential at any point is negative if q is negative. 2. The potential at any point is positive if q is positive. 3. The potential is zero at r = infinity. To find the electric potential V of a severe point charges, express as It is the scalar sum of the fields produced by each charge.
• 18.
  A point chargehas a charge of 8.00 x 10-11 C. At what distance from the point charge is the electrical potential 12.0 V? 24.0 V? given unknown sketch Q1 = +8.00 x 10-11 C ; V1 = 12.0 V ; V2 = 24.0 V r
• 19.
  A point chargehas a charge of 8.00 x 10-11 C. At what distance from the point charge is the electrical potential 12.0 V? 24.0 V? strategy solution Re-arrange the formula and solve for r since it is unknown, then plug-in the values. 𝑟 = 1 4𝜋𝜖0 𝑞 𝑉 r = 1 4𝜋𝜖0 (8.00 𝑥10−11𝐶) 24 𝑉 r = 9 x 109 𝑁𝑚2 𝐶2 3.33 𝑥 10−12𝐶 𝐽/𝐶 r = 0.03 m at 24 V
• 20.
  A point chargehas a charge of 8.00 x 10-11 C. At what distance from the point charge is the electrical potential 12.0 V? 24.0 V? solution 𝑟 = 1 4𝜋𝜖0 𝑞 𝑉 r = 1 4𝜋𝜖0 (8.00 𝑥10−11𝐶) 12 𝑉 r = 9 x 109 𝑁𝑚2 𝐶2 6.67 𝑥 10−12𝐶 𝐽/𝐶 r = 0.06 m at 12 V
• 21.
  DO THIS!!! In a certainregion an electric potential V is present. An unknown charge Q is moved around this region between points at potential difference ∆V. A physicist measures the change in the potential energy ∆U as the charge is moved. The table shows the data of the measurements. The unit for ∆V are MV (mega-Volts), the units for ∆U are kJ (kilo- joules). ∆𝐕 52.1 24.9 11.7 4.97 3.55 3.32 1.52 1.87 ∆𝐔 357.8 152.6 119.3 43.8 48.9 43.8 32.9 42.3