ELECTRIC POTENTIAL ALI ALZAHRANI
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ELECTRIC POTENTIAL ALI ALZAHRANI ELECTRIC POTENTIAL ο½ Electric potential energy is the energy needed to change the spatial configuration of charged particles from initial to final destinations. βπ = ππ − ππ = −π ο½ We used to take the energy at infinity is zero, therefore, the energy required to move any charged particle to any position is π = −π ο½ Electric potential ‘potential’ is defined as the potential energy per unit charge, such as π= π π ELECTRIC POTENTIAL ο½ The potential difference between any two points is defined as βπ = ππ − ππ = ο½ ππ ππ βπ π − = =− π π π π The SI unit of potential is ‘volt (V)’ which is also J/C. However, the potential energy is measured by ‘electron-volt (eV)’ which is 1.6 × 10−19 J. 1 eV = 1.6× 10−19 C. (1 J/C) = 1.6× 10−19 J ο½ Equipotential surfaces are those where each point on them have the same electric potential or the potential difference between any two points is zero. Therefore no net work should be done on a charged particle to move between any two points on that surface. EQUIPOTENTIAL SURFACE - EXAMPLES (a) A single positive charge (c) Two positive charges (b) An electric diploe (d) Uniform electric field ELECTRIC POTENTIAL AND FIELD ο½ The electric potential is defined in terms of electric field as π βπ = − πΈ. ππ π From this we note that 1 V = 1 N.m/C or 1 N/C = 1 V/m. If the initial potential is zero, the above equation becomes π π=− πΈ. ππ π NOTE: It is important to note that the potential is a scalar quantity NOT a vector. ELECTRIC POTENTIAL DUE TO A POINT CHARGE ο½ The electric potential of a point charge π at any distance π can be evaluated from π π=− πΈ. ππ π Having known that the electric field due to any point charge is πΈ = π π=− ππ ππ ππ = 2 π ∞ π NOTE: Positive charges produce positive potential and vice versa. ππ π2 we get ELECTRIC POTENTIAL DUE TO A GROUP OF POINT CHARGES ο½ The electric potential of a collection of point charge ππ at any distance ππ is the summation of all potentials of each individual point charge π π= π ππ = π π=1 π=1 ππ ππ ο½ The electric potential due to an electric dipole π at any point at distance π and makes an angle π with the axes of the dipole is defined as π π cos π π2 NOTE: The angle π is measured from the direction of the dipole. π= ELECTRIC POTENTIAL DUE TO A CONDUCTING SPHERE ο½ The electric potential of a conducting sphere of radius π and charge π at any distance π outside the sphere, where the electric field (outside) is πΈ = ππ π2 , is π π=− ο½ ππ ππ ππ = 2 π ∞ π The electric potential of an insulating sphere of radius π and charge π at any distance π inside the sphere, where the electric field (outside) is πΈ = π π=− ∞ ππ π2 ππ ππ − π2 and (inside) πΈ = 0, is π 0 ππ = π ππ π Therefore the electric potential at any point of the isolated conductor is invariant, leading to that no extra work is needed to carry a charge from surface to any point within the conductor, see example 7. ELECTRIC POTENTIAL DUE TO AN INSULATING SPHERE ο½ The electric potential of an insulating sphere of radius π and charge π at any distance π outside the sphere, where the electric field (outside) is πΈ = ππ π2 , is π ππ ππ ππ = 2 π ∞ π π=− ο½ The electric potential of a conducting sphere or radius π and charge π at any distance π inside the sphere, where the electric field (outside) is πΈ = π π=− ∞ ππ ππ − π2 π π ππ π2 and (inside) πΈ = ππ π ππ π2 − π 2 ππ = 1− π 3 π 2π 2 ππ π , π 3 is POTENTIAL ENERGY OF A SYSTEM OF POINT CHARGES ο½ The electric potential energy of a system of point charges ππ separated by distances ππ is the summation of all potential energy of these point charges π π= π πππ = π π,π=1 π,π=1 where πππ is the distance between charges ππ and ππ ππ ππ πππ E DERIVED FROM V ο½ The electric field can be evaluated from the expression of the potential such as πΈ=− ∂π ∂π In our three-dimensional space, the above definition can be rewritten as πΈπ₯ = − ππ ππ₯ πΈπ¦ = − ππ ππ¦ Therefore the electric field vector is πΈ=− ππ ππ ππ π− π− π ππ₯ ππ¦ ππ§ πΈπ§ = − ππ ππ§ 1. Calculate the electric potential at a distance 5 cm from a point charge 2.5 nC. Solution The magnitude of the electric potential at any distance from a point charge is given by π= ππ π Hence the electric potential at 5 cm from the charge is π= ππ π = 9×109 ×2.5×10−9 0.05 = 450 V 2. A point charge produces an electric field of 180 N/C at 2 cm. Calculate the electric potential at 4 cm from the charge. Solution The magnitude of the electric field due to a point charge is given by πΈ= π π π2 Therefore to find the electric potential at any point, we have to determine the charge, firstly. π= πΈ π2 π = 180×0.022 9×109 = 8 × 10−12 C Hence the electric potential at 4 cm from the charge is π= π π 9 × 109 × 8 × 10−12 = = 1.8 V π 0.04 3. As shown in the arrangement below, calculate the electric potential at point A. Calculate the work needed to bring a charge Q=6 nC from infinity to point A. Solution A ο½ 2m 3 µC 3m -4 µC The electric potential of a collection of points is π = π1 + π2 = π π1 π π2 π1 π2 3 × 10−6 −4 × 10−6 + =π + = 9 × 109 × + = 6300 V π1 π2 π1 π2 2 5 Therefore the work needed to bring a charge from infinity to this point is π = −ππ = −6 × 10−9 × 6300 = −3.78 × 10−5 J = −2.36 × 1014 eV 4. Four charges 1, 2, 3 and -4 µC are located at the corners of a square of side 3 m as shown in the figure. Calculate the electric potential energy of the system. Solution q2=2 µC q3=3 µC q1=1 µC q4=-4 µC The potential energy is ππ1 π2 ππ2 π3 ππ3 π4 ππ4 π1 π 9 × 109 π= + + + = π1 π2 + π2 π3 + π3 π4 + π4 π1 = 2 + 6 − 12 − 4 × 10−12 = −24 mJ π12 π23 π34 π41 π 3 5. Two equal and opposite charges 6.0 µC and -6.0 µC are separated by a distance of 2 cm. What is the electric potential at a point 30 cm away from the axis of the dipole? Solution The electric potential due to a dipole is π= π π cos π π2 However along the axis of the dipole the angle is zero, (cos π = 1), therefore π= π π π ππ 9 × 109 × 6 × 10−6 × 0.02 = 2 = = 12 kV π2 π 0.32 6. A metallic sphere of radius 10 cm has a charge of 5.0 µC. What is the electric potential at (i) 5 cm and (ii) 15 cm from its center? Solution (i) The electric potential at any point inside a conducting sphere is π= π π 9 × 109 × 5 × 10−6 = = 450000 V π 0.1 (ii) The electric potential at any point outside a conducting sphere is π π 9 × 109 × 5 × 10−6 π= = = 300000 V π 0.15 7. A conducting sphere of radius 10 cm has a charge of 5.0 µC. Calculate the work required to carry a charge 2.0 µC from the surface to the center of the sphere. Solution We know that the electric potential at any point inside a conducting sphere is π= ππ π π= ππ π Also the electric potential at the surface is Therefore the potential difference is zero, leading the work is zero, too, according to π = −π βπ 8. At a certain region, the electric potential is found to be π½ π, π, π = πππ π + π. Calculate the electric field at point (1,0,1). Solution The electric field relates to potential through πΈπ₯ = − ππ = −6π₯π¦ ππ₯ πΈπ¦ = − ππ = −3π₯ 2 ππ¦ πΈπ§ = − Therefore at the point (1,0,1) we get πΈπ₯ = 0 πΈπ¦ = −3 Hence the electric field is πΈ = −3π − π πΈπ§ = −1 ππ = −1 ππ§
