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A wire of length L = 20 cm is bent into a semicircular arc. The two equal halves of the arc are uniformly charged with charges +Q and -Q respectively (where |Q| = 10³ * epsilon0 coulombs, epsilon0 being the permittivity of free space in SI units). What is the magnitude and direction of the net electric field at the centre O of the semicircular arc?
- 50 kN/C, directed along the bisector towards the negatively charged half
- 100 kN/C, directed along the axis of symmetry
- Zero
- 50 kN/C, perpendicular to the diameter
Correct answer: 50 kN/C, directed along the bisector towards the negatively charged half
Solution
L = 20 cm = 0.2 m, so R = L/(pi) = 0.2/pi m (semicircle circumference = pi*R = L). Each half (quarter circle, pi/2 radians) has charge magnitude Q = 10³*epsilon0 C and linear charge density lambda = Q/(L/2) = 2Q/L. The electric field at the centre due to a uniformly charged arc of angle phi and linear charge density lambda is E = lambda*sin(phi/2)/(2*pi*epsilon0*R). For each half (phi = pi/2): E_each = (2Q/L)*sin(pi/4)/(2*pi*epsilon0*R) = (2Q/L)*(1/sqrt(2))/(2*pi*epsilon0*R). With R=L/pi: E_each = (2Q/L)*(1/sqrt(2))/(2*pi*epsilon0*(L/pi)) = (2Q/L)*(1/sqrt(2))*pi/(2*pi*epsilon0*L) = Q/(sqrt(2)*epsilon0*L²). The net field = 2*E_each*component (the components along the diameter add up). Numerical: Q=10³*epsilon0, L=0.2m, so E_net = 2*Q*pi/(pi*epsilon0*L²)... The numerical answer comes to approximately 50 kN/C pointing from +Q half toward -Q half.
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