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- 19:26 ☺ T 1.5. Study the following problem that was solved directly from Coulomb's law and answer the set of questions that follow. . Problem Calculate the electric field at a point P created by a thin, long, straight filament, electrically charged with a constant linear density A, using the direct approach from Coulomb's law. Solution () From the translational and axial symmetries of the problem, it is clear that E(r) = nE(p), where p is the shortest distance from the observation point to the filament". Let us select the plane of drawing so that it contains both the filament and the observation point, and take the line of the filament for axis : (see the figure below) dE cos No² +1²y²³² Then, according to the linear superposition principle, the field's magnitude may be calculated as Elp) - )= dE, - dE cos 0= - E- +222 where dE is the magnitude of the elementary contribution to the field, created by a small segment dz of the filament, with electric charge Adz. But 1 1 dE=Adz- Anco p²…arrow_forwardQuestion 1 a) In J. J. Thomson experiment (1897), an electron moving horizontally with a constant speed vo enters in between the horizontal plates of a capacitor. The electric field strength between the plates of length L and distance d, is E. The vertical deviation of the electron at the moment of exit from the field region is measured to be Y. Derive the expression giving the electron's charge to mass ratio, i.e. e/m to be 2v,Y/CEL). (Recall that Thomson received Nobel Prize for his achievement.) b) Calculate e/m, knowing the following data. E=1.6x10* Newton/Coulomb, L=10 cm, Y=2.9 cm, v=2.19x10* km/s. (Be careful to use coherent units.)arrow_forwardE = E1 E = 0 E= E1 E= E1 The electric field is measured all over a cubical surface, and the pattern of field detected is shown in the figure above. On the right side of the cube, the electric field has magnitude Ej = 426 V/m, and the angle between the electric field and the surface of the cube is 0 = 15 degrees. On the bottom of the cube, the electric field has the same magnitude Ej, and the angle between the electric field and the surface of the cube is also 0 = 15 degrees. On the top of the cube and the left side of the cube, the electric field is zero. On half of the front and back faces, the electric field has magnitude Ej and is parallel to the face; on the other half of the front and back faces, the electric field is zero. One edge of the cube is 41 cm long. Part 1 What is the net electric flux on this cubical surface? Net electric flux = i V•m Save for Later Attempts: 0 of 4 used Submit Answerarrow_forward
- Useful Constants: k = 9.00 × 10º Nm² C2 €0 = 8.85 × 10–12 C² Nm2 e = 1.6 × 10¬19C %3D mẹ = 9.11 x 10~31 kg %3D 27 kg = 1.67 × 10 mp mn = 1.68 × 10-27kgarrow_forwardAn electron is released 8.7 cm from a very long nonconducting rod with a uniform 7.0 μC/m. What is the magnitude of the electron's initial acceleration?arrow_forwardIn Thomson’s model, an atom is a positively charged spherical material in which negatively charged electrons are embedded like chocolate chips on a ball of cookie dough. Consider such an atom, made up of a uniformly charged sphere with charge +e and radius R and a point charge with mass m and charge −e. a. Locate the position of electrostatic equilibrium for the electron inside the sphere. b. Assume further that the sphere has little or no resistance to the electron’s mo- tion. If the electron is displaced from equilibrium by a distance less than R, show that the resulting motion of the electron would be simple harmonic. c. If the electron was displaced from equilibrium by a distance greater than R, would the electron oscillate? Would its motion be simple harmonic?arrow_forward
- Suppose a capacitor consists of two coaxial thin cylindrical conductors. The inner cylinder of radius ra has a charge of +Q, while the outer cylinder of radius rp has charge -Q. The electric field E at a radial distance r from the central axis is given by the function: E = ae-r/ao + B/r + bo where alpha (a), beta (B), ao and bo are constants. Find an expression for its capacitance. First, let us derive the potential difference Vab between the two conductors. The potential difference is related to the electric field by: ['´e Vob = Edr= - Edr Calculating the antiderivative or indefinite integral, Vab = (-aaoe¯r7ao + B + bo By definition, the capacitance C is related to the charge and potential difference by: C = Evaluating with the upper and lower limits of integration for Vab, then simplifying: C = Q / ( (erb/ao - eralao) + B In( ) + bo ( ))arrow_forwardDon't use chat gpt It Chatgpt means downvotearrow_forwardSuppose a capacitor consists of two coaxial thin cylindrical conductors. The inner cylinder of radius ra has a charge of +Q, while the outer cylinder of radius rh has charge -Q. The electric field E at a radial distancer from the central axis is given by the function: E = ae-r/ao + B/r + bo where alpha (a), beta (B), ao and bo are constants. Find an expression for its capacitance. First, let us derive the potential difference Vab between the two conductors. The potential difference is related to the electric field by: Vab = Edr = - Edr Calculating the antiderivative or indefinite integral, Vab = (-aaoe-r/ao + B + bo By definition, the capacitance C is related to the charge and potential difference by: C= Q I Vabarrow_forward
- Calculate the magnitude of strength of an electric field at the distance of 12.0 nm from the nucleus of the Uranium-235 atom?arrow_forward04 Two very small spheres are initially neutral and separated by a distance of $6 cm. Suppose that 310¹1 electrons are removed from one sphere and placed on the other. (a) What is the magnitude of the electrostatic force that acts on cach sphere? (b) is the force attractive or repulsive? Why?arrow_forwardSuppose a capacitor consists of two coaxial thin cylindrical conductors. The inner cylinder of radius ra has a charge of +Q, while the outer cylinder of radius rb has charge -Q. The electric field E at a radial distance r from the central axis is given by the function: E = αe-r/a0 + β/r + b0 where alpha (α), beta (β), a0 and b0 are constants. Find an expression for its capacitance. First, let us derive the potential difference Vab between the two conductors. The potential difference is related to the electric field by:arrow_forward
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