The electric field at the point x = 5.00cm and y = 0points in the positive x dir

Question

The electric field at the point x = 5.00cm and y = 0points in the positive x direction with a magnitude of 8.00 N/C. Atthe point x = 10.0cm and y = 0 the electric field points in thepositive x direction with a magnitude of 15.0 N/C. Assume thiselectric field is produced by a single point charge. A.) Find the charge's location. x = B.) Find the sign of the charge. C.) Find the magnitude of the charge. q = The electric field at the point x = 5.00cm and y = 0points in the positive x direction with a magnitude of 8.00 N/C. Atthe point x = 10.0cm and y = 0 the electric field points in thepositive x direction with a magnitude of 15.0 N/C. Assume thiselectric field is produced by a single point charge. A.) Find the charge's location. Find the charge's location. x = B.) Find the sign of the charge. C.) Find the magnitude of the charge. Find the magnitude of the charge. q =

Explanation / Answer

Let the charge be q Electric field at x= 10 cm is high than electric field at x =5 cm .So, the near to x = 10 cm Distance of the point charge from x = 10 is r electric field at x = 10 cm is E = 15 N / C                                       K q / r ^ 2 = 15 N / C where K = coulomb's constant = 8.99 * 10 ^ 9 N m^2 / C ^2 So, q / r ^ 2 = 15 / ( 8.99*10^9 )                    = 1.6685 * 10 ^ -9                    q = 1.6685* 10 ^ -9 * r ^ 2     ---( 1) electric field at x = 5 cm is E ' = 8 N / C                                  Kq / ( r+ 5 cm )^ 2 = 8                                   q / ( r+ 5 ) ^ 2 = 0.8898 * 10 ^ -9                                                         q = 0.8898* 10 ^ -9 * ( r + 5 )^ 2 ---( 2) from eq ( 1 ) and ( 2 ) , 1.6685* 10 ^ -9 * r ^ 2 =0.8898* 10 ^ -9 * ( r + 5 )^ 2                                                                   r ^ 2 = 0.53329 ( r + 5 ) ^ 2                                                                         = 0.53329 ( r ^ 2 + 10 r + 25 ) 0.4667 r ^ 2 -5.3329 r + 13.3322 = 0 solve this quadratic equation we get r value So, Position of charge = 10 cm + r The charge is negative

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