t View History Bookmarks People Window Help x Concordia University PHY X M Denis

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t View History Bookmarks People Window Help x Concordia University PHY X M Denise nanny position den x THIS CHEERED HER UPI VL x c Ch inglearning.com/ibiscms/modlibis/view.php?ida3196117 score 1/20/2017 09:00 AM 55.1/100 G 01:49 AM 1/20/2017 Me t Print Calculator Periodic Table Question 10 of 19 A Sapling Learning An electron and a proton are fixed at a separation distance of 835 mm. Find the magnitude and direction of the electric field at their midpoint. NIC O Toward the proton O cannot be determined O Toward the electron Perpendicular to the line of the particles

Explanation / Answer

EF due to electron = kq/r^2

= 9*10^9*1.6*10^-19 / (835/2 * 10^-9)^2

= 8261.32 N/C (towards the electron)

EF due to proton =kq/r^2

= 9*10^9*1.6*10^-19 / (835/2 * 10^-9)^2

= 8261.32 N/C (towards the electron)

Net EF = 8261.32 + 8261.32 = 16522.64 N /C towards the electron

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