Electric potential difference ppt

Field lines and equipotentials and electric potential and potential difference ppt and electric current and potential difference ppt
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Dr.TomHunt,United States,Teacher
Published Date:23-07-2017
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Electric Potential II Physics 2415 Lecture 7 Michael Fowler, UVa Today’s Topics • Field lines and equipotentials • Partial derivatives • Potential along a line from two charges • Electric breakdown of air Potential Energies Just Add • Suppose you want to bring one • a y charge Q close to two other fixed Q r 1 charges: Q and Q . 1 2 Q 1 r • The electric field Q feels is the r sum of the two fields from Q , Q , 2 1 2 the work done in moving d is Q Ed Ed Ed 2 12 so since the potential energy 0 x change along a path is work done,  1 QQ 12 Vr   V r V r V r  12 4  rr 0 1 2Total Potential Energy: Just Add Pairs • If we begin with three charges Q , 1 • a Q and Q initially far apart from 2 3 Q each other, and bring them closer 1 Q r 3 13 together, the work done—the potential energy stored—is  1 QQ Q Q Q Q 12 2 3 3 1 r r U 23  12 4  r r r 0 12 23 31 and the same formula works for Q assembling any number of 2 charges, just add the PE’s from all pairs—avoiding double counting Equipotentials • Gravitational equipotentials are just contour lines: lines connecting points (x,y) at the same height. (Remember PE = mgh.) • It takes no work against gravity to move along a contour line. • Question: What is the significance of contour lines crowding together? Electric Equipotentials: Point Charge • The potential from a point charge Q is 1 Q Vr  4  r 0 • Obviously, equipotentials are surfaces of constant r: that is, spheres centered at the charge. • In fact, this is also true for gravitation—the map contour lines represent where these spheres meet the Earth’s surface. Plotting Equipotentials • Equipotentials are surfaces in three dimensional space—we can’t draw them very well. We have to settle for a two dimensional slice. • Check out the representations here. Plotting Equipotentials • . Here’s a more physical representation of the electric potential as a function of position described by the equipotentials on the right. Given the Potential, What’s the Field? • Suppose we’re told that some static charge distribution gives rise to an electric field V x,, y z corresponding to a given potential .  E x,, y z  • How do we find ? • We do it one component at a time: for us to x,, y z x  x,, y z push a unit charge from to   Ex takes work , and increases the PE of the x charge by . V xx, y, zV x, y, z  • So: V xx, y, zV x, y, zV x, y, z  Ex for  0. x  xxWhat’s a Partial Derivative? • The derivative of f(x) measures how much f changes in response to a small change in x. • It is just the ratio f/x, taken in the limit of small x, and written df/dx. • The potential function is a function of V x,, y z  three variables—if we change x by a small amount, keeping y and z constant, that’s partial differentiation, and that measures the field component in the x direction: V x, y, zV x, y, zV x, y, z  E , E , E . x y z xyzField Lines and Equipotentials • The work needed to move unit charge a tiny distance at position is . d  E r d r  • That is, V r dV rE rd   • Now, if d is pointing along an equipotential, by definition V doesn’t change at all • Therefore, the electric field vector Er at any  point is always perpendicular to the equipotential surface. Potential along Line of Centers of Two Equal Positive Charges V(x) • D x 0 Q Q Note: the origin (at the midpoint) is a “saddle point” in a 2D graph of the potential: a high pass between two hills. It slopes downwards on going away from the origin in the y or z directions. Potential along Line of Centers of Two Equal Positive Charges V(x) x 0 Q Q • Clicker Question: • At the origin in the graph, the electric field E is: x A. maximum (on the line between the charges) B. minimum (on the line between the charges) C. zero Potential along Line of Centers of Two Equal Positive Charges V(x) x 0 Q Q • Clicker Answer: V E  E (0) = Zero: because equals minus the slope. x x x • (And of course the two charges exert equal and opposite repulsive forces on a test charge at that point.) Potential and field from equal +ve charges • . • . Potential along Bisector Line of Two Equal Positive Charges Q V(y) a r y 0 a Now plotting potential along the y-axis, not the x-axis Q • For charges Q at y = 0, x = a and x = -a, the potential at a point on the y-axis: 22 kQ kQ Vy  22 r ay Note: same formula will work on axis for a ring of charge, 2Q becomes total charge, a radius. Potential from a short line of charge • Rod of length 2 has uniform charge density , 2 = Q. What is • . the potential at a point P in the bisector plane? • The potential at y from the charge P between x, x +  x is kQ k  x k x x r y  22 rr xy • So the total potential x 22  y kdx kQ Vy ln   2 2 2 2 2 x y y  Great – but what does V(y) look like? Potential from a short line of charge 22  y kdx kQ Vy ln   • . 2 2 2 2 2 x y y  • What does this look like at a large y distance ? y • Useful math approximations: for 1 1 x 1 x, ln 1 x x  small x, • So 22 1/ y  y y  1 2 / y  22 yy 1/   y x kQ kQ V y ln 1 2 / y • And  2 y Bottom line: at distances large compared with the size of the line, it looks like a point charge. Potential from a long line of charge • Let’s take a conducting cylinder, radius R. • If the charge per unit length of cylinder is , the external electric field points radially outwards, from symmetry, and has magnitude E(r) = 2k/r, from Gauss’s theorem. rr  dr  V r V R E rdr V R 2k • So    r RR  V R 2k ln r ln R .  • Notice that for an infinitely long wire, the potential keeps on increasing with r for ever: we can’t set it to zero at infinity Potential along Line of Centers of Two Equal but Opposite Charges V(x) • D x 0 Q -Q