Two charged particles on an x-axis: -q1 = -3.2 x 10 ^ -19C at x= -5.00m and q2 =
ID: 1759102 • Letter: T
Question
Two charged particles on an x-axis: -q1 = -3.2 x 10 ^ -19C at x= -5.00m and q2 = 3.20 x 10 ^-19 at x = +5.00m.. What are the (a) magnitude and (b) direction (relative to the positive direction of the x axis) of the netelectric field produced at point P at y = 10m ... Two charged particles on an x-axis: -q1 = -3.2 x 10 ^ -19C at x= -5.00m and q2 = 3.20 x 10 ^-19 at x = +5.00m.. What are the (a) magnitude and (b) direction (relative to the positive direction of the x axis) of the netelectric field produced at point P at y = 10m ...Explanation / Answer
Given -q= -3.2 x 10 ^ -19 Cq = 3.20 x 10 ^ -19 C x = -5 m x ' = 5 m y = 10 m X-component of Electric filed at point P due to charge q is E = K q / x ^ 2 where K = coulomb's constant = 8.99 * 10 ^ 9 N m ^ 2 / C^ 2 pulg the values we get E value X-component of Electric filed at point P due to charge q' is E ' = K q / x ' ^ 2 where K = coulomb's constant = 8.99 * 10 ^ 9 N m ^ 2 / C^ 2 pulg the values we get E ' value Y-component of Electric filed at point P due to charge qis E " = K q / y ^ 2 where K = coulomb's constant = 8.99 * 10 ^ 9 N m ^ 2 / C^ 2 pulg the values we get E " value Y-component of Electric filed at point P due to chargeq ' is E = K q / y ^ 2 where K = coulomb's constant = 8.99 * 10 ^ 9 N m ^ 2 / C^ 2 pulg the values we get E "' value Resulatnt X-component of electric field Ex = E + E" Resultant Y-component of electric field Ey = E ' + E"' Therefore Magnitude of resultant electric field =[ Ex ^ 2 + E y ^ 2 ] plug the values we get answer Let resulatnt electric field makes an angle with X-axisthen tan = Ey / Ex = tan -1 ( Ey / Ex ) X-component of Electric filed at point P due to charge q' is E ' = K q / x ' ^ 2 where K = coulomb's constant = 8.99 * 10 ^ 9 N m ^ 2 / C^ 2 pulg the values we get E ' value Y-component of Electric filed at point P due to charge qis E " = K q / y ^ 2 where K = coulomb's constant = 8.99 * 10 ^ 9 N m ^ 2 / C^ 2 pulg the values we get E " value Y-component of Electric filed at point P due to chargeq ' is E = K q / y ^ 2 where K = coulomb's constant = 8.99 * 10 ^ 9 N m ^ 2 / C^ 2 pulg the values we get E "' value Resulatnt X-component of electric field Ex = E + E" Resultant Y-component of electric field Ey = E ' + E"' Therefore Magnitude of resultant electric field =[ Ex ^ 2 + E y ^ 2 ] plug the values we get answer Let resulatnt electric field makes an angle with X-axisthen tan = Ey / Ex = tan -1 ( Ey / Ex ) Y-component of Electric filed at point P due to charge qis E " = K q / y ^ 2 where K = coulomb's constant = 8.99 * 10 ^ 9 N m ^ 2 / C^ 2 pulg the values we get E " value Y-component of Electric filed at point P due to chargeq ' is E = K q / y ^ 2 where K = coulomb's constant = 8.99 * 10 ^ 9 N m ^ 2 / C^ 2 pulg the values we get E "' value Resulatnt X-component of electric field Ex = E + E" Resultant Y-component of electric field Ey = E ' + E"' Therefore Magnitude of resultant electric field =[ Ex ^ 2 + E y ^ 2 ] plug the values we get answer Let resulatnt electric field makes an angle with X-axisthen tan = Ey / Ex = tan -1 ( Ey / Ex ) Y-component of Electric filed at point P due to chargeq ' is E = K q / y ^ 2 where K = coulomb's constant = 8.99 * 10 ^ 9 N m ^ 2 / C^ 2 pulg the values we get E "' value Resulatnt X-component of electric field Ex = E + E" Resultant Y-component of electric field Ey = E ' + E"' Therefore Magnitude of resultant electric field =[ Ex ^ 2 + E y ^ 2 ] plug the values we get answer Let resulatnt electric field makes an angle with X-axisthen tan = Ey / Ex = tan -1 ( Ey / Ex )
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