V q.. REASONING The potential V created by a point charge q at a spot that is located at a

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Dora Stafford
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1 8. REASONING The electric potential at a istance r from a point charge q is given by Equation 9.6 as kq / r. The total electric potential at location P ue to the four point charges is the algebraic sum of the iniviual potentials. q q q SOUTION The total electric potential at P is (see the rawing) k q k q k q k q kq q P Substituting in the numbers gives 9 Nm kq m. REASONING The potential create by a point charge q at a spot that is locate at a kq istance r is given by Equation 9.6 as, where q can be either a positive or negative r quantity, epening on the nature of the charge. We will apply this expression to obtain the potential create at the empty corner by each of the three charges fixe to the square. The total potential at the empty corner is the sum of these three contributions. Setting this sum equal to zero will allow us to obtain the unknown charge. SOUTION The rawing at the right shows the three charges fixe to the corners of the square. The length of each sie of the square is enote by. Note that the istance +.8 between the unknown charge q an the empty corner is. Note also that the istance between one of the.8-μ charges an the empty corner is r, but that the istance between the other.8-μ charge an the empty corner is +.8 r, accoring to the Pythagorean theorem. q Using Equation 9.6 to express the potential create by the unknown charge q an by each of the.8-μ charges, we fin that the total potential at the empty corner is total 6.80 k.80 6 kq k 0 In this result the constant k an the length can be eliminate algebraically, leaing to the following result for q:

2 q.80 0 or q REASONING The net work W AB one by the electric force on the point charge q 0 as it moves from A to B is proportional to the potential ifference B A between those positions, accoring to W AB B A (Equation 9.4). Positions A an B are on ifferent q0 equipotential surfaces, so we will rea the potentials B an A from the rawing. We will employ the same proceure to solve part (b). SOUTION a. Solving W AB B A (Equation 9.4) for W AB, we obtain q0 W q () AB B A 0 From the rawing, we see that A = an B = Therefore, from Equation (), W AB J b. Positions A an are both on the equipotential surface. Aapting Equation (), then, we obtain 7 WA q0 A J 38. REASONING The electric force F is a conservative force, so the total energy (kinetic energy plus electric potential energy) remains constant as the electron moves across the capacitor. Thus, as the electron accelerates an its kinetic energy increases, its electric potential energy ecreases. Accoring to Equation 9.4, the change in the electron s electric potential energy is equal to the charge on the electron (e) times the potential ifference between the plates, or EPE positive EPE negative = (e)( positive negative ) (9.4)

3 The electric fiel E is relate to the potential ifference between the plates an the isplacement s by E positive negative (Equation 9.7a). Note that s an s ( positive negative ) are positive numbers, so the electric fiel is a negative number, enoting that it points to the left in the rawing: Electric fiel + Electron F + + x +x + SOUTION The total energy of the electron is conserve, so its total energy at the positive plate is equal to its total energy at the negative plate: KE EPE KE EPE positive positive negative negative Total energy at positive plate Total energy at negative plate Since the electron starts from rest at the negative plate, KE negative = 0 J. Thus, the kinetic energy of the electron at the positive plate is KE positive = (EPE positive EPE negative ). We know from Equation 9.4 in the REASONING section that EPE positive EPE negative = (e)( positive negative ), so the kinetic energy can be written as KE positive = (EPE positive EPE negative ) = e ( positive negative ) Since the potential ifference is relate to the electric fiel E an the isplacement s by E s (Equation 9.7a), we have that positive negative apacitor plates KE positive = e positive negative = e( E s) = /m0.0 m 4.00 J In arriving at this result, we have use the fact that the electric fiel is negative, since it points to the left in the rawing.

4 46. REASONING Equation 9.0 gives the capacitance as = 0 A/, where is the ielectric constant, an A an are, respectively, the plate area an separation. Other things being equal, the capacitor with the larger plate area has the greater capacitance. The iameter of the circle equals the length of a sie of the square, so the circle fits within the square. The square, therefore, has the larger area, an the capacitor with the square plates woul have the greater capacitance. To make the capacitors have equal capacitances, the ielectric constant must compensate for the larger area of the square plates. Therefore, since capacitance is proportional to the ielectric constant, the capacitor with square plates must contain a ielectric material with a ielectric constant. Thus, the capacitor with circular plates contains the material with the greater ielectric constant. SOUTION The area of the circular plates is A = circle, while the area of the square plates is A square =. Using these areas an applying Equation 9.0 to each capacitor gives circle0 an Since the values for are the same, we have circle0 circle square 0 square 0 4 or circle 4 square square 48. REASONING The energy store in a capacitor is given by Energy (Equation 9.b), where is the capacitance of the capacitor an is the potential ifference across its plates. The only ifference between the two capacitors is the ielectric material (ielectric constant = 4.50) insie the fille capacitor. Therefore, the fille capacitor s capacitance is greater than the capacitance of the empty capacitor by a factor of : ()

5 Because both capacitors store the same amount of energy, from Energy (Equation 9.b), we have that () where is the potential ifference across the plates of the fille capacitor, an =.0 is the potential ifference across the plates of the empty capacitor. SOUTION Solving Equation () for, we obtain or Substituting Equation () into Equation (3) yiels (3) REASONING The charge q store on the plates of a capacitor connecte to a battery of voltage is q (Equation 9.8). The capacitance is 0 A (Equation 9.0), where κ is the ielectric constant of the material between the plates, ε 0 is the permittivity of free space, A is the area of each plate, an is the istance between the plates. Once the capacitor is charge an isconnecte from the battery, there is no way for the charge on the plates to change. Therefore, as the istance between the plates is ouble, the charge q must remain constant. However, Equation 9.0 inicates that the capacitance is inversely proportional to the istance, so the capacitance ecreases as the istance increases. In Equation 9.8, as ecreases, the voltage must increase in orer that q remains constant. The voltage increases as a result of the work one in moving the plates farther apart. In solving this problem, we will apply Equations 9.8 an 9.0 to the capacitor twice, once with the an once with the larger value of the istance between the plates. SOUTION Using q (Equation 9.8) an 0 A (Equation 9.0), we can express the charge on the capacitor as follows: 0A 0A q

6 where we have mae use of the fact that, since the capacitor is empty. Applying this result to the capacitor with an larger values of the istance, we have 0A A q an q 0 larger larger Since q is the same in each of these expressions, it follows that A A or 0 0 larger larger larger larger Thus, we fin that the voltage increases to a value of larger larger 9.0 8

1. The electron volt is a measure of (A) charge (B) energy (C) impulse (D) momentum (E) velocity

AP Physics Multiple Choice Practice Electrostatics 1. The electron volt is a measure of (A) charge (B) energy (C) impulse (D) momentum (E) velocity. A soli conucting sphere is given a positive charge Q.