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\nThis question relates to Coulomb’s Law, which gives the force per unit charge due to the interaction between stationary charges.
\nWhen providing numerical answers you may express them using scientific notation. Express values to four significant figures and use the values of physical constants as provided in the course notes, and take the relative permittivity of air to be 1.0005.
\nIf you got the wrong answer, did you remember to
\nThe $E$-field due to a point charge is
\n$\\displaystyle{q\\over 4 \\pi \\varepsilon_0 \\varepsilon_r r^2}$
\nradially away from the point charge. The force experienced in an electric field is generally $\\vec{F}=q\\vec{E}$, so the force on a point charge, $q_1$ due to another point charge, $q_2$ has a magnitude of
\n$\\displaystyle q_1 {q_2\\over 4 \\pi \\varepsilon_0 \\varepsilon_r r^2}$.
\nSince typically $\\varepsilon_r\\ge1$, any material between the two charges tends to reduce the force relative to the point charges in vacuum. It is important to note that the question asks for the percentage by which the force is reduced, not the percentage of the original force that remains.
\nThe force due to gravity is $mg$ towards the centre of the Earth, so one can calculate the force due the electric field and work out the mass that when mulitplied by $g$ has the same magnitude.
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", "templateType": "anything"}, "eps0": {"name": "eps0", "group": "Ungrouped variables", "definition": "8.854*10^-12", "description": "Permittivity of free space in F/m.
", "templateType": "anything"}, "mass": {"name": "mass", "group": "Ungrouped variables", "definition": "force1/g", "description": "Mass that would balance the force between the charges on Earth (kg).
", "templateType": "anything"}, "r1": {"name": "r1", "group": "Ungrouped variables", "definition": "random(0.5..2.5#0.5)", "description": "Distance of the second point charge from the first in cm.
", "templateType": "anything"}, "epsmat": {"name": "epsmat", "group": "Ungrouped variables", "definition": "random(4..10#0.1)", "description": "Randomised relative permittivity of material substituting for air (no units)
", "templateType": "anything"}, "q2": {"name": "q2", "group": "Ungrouped variables", "definition": "siground(random(0.5..2.5#0.5 except q1),2)", "description": "Magnitude of charge for which the electric field is originating in $\\mu$C, randomised in steps of 0.5$\\mu$C from 0.5 to 2.5$\\mu$C, but different from $q_1$.
", "templateType": "anything"}, "epsair": {"name": "epsair", "group": "Ungrouped variables", "definition": "1.0005", "description": "Relative permittivity of air (dimensionless quantity).
", "templateType": "anything"}, "force2": {"name": "force2", "group": "Ungrouped variables", "definition": "q1*q2*(10^(-12))/(4*pi*eps0*epsmat*(r1*(10^(-2)))^2)", "description": "", "templateType": "anything"}, "q1": {"name": "q1", "group": "Ungrouped variables", "definition": "siground(random(0.5..2.5#0.5),2)", "description": "Magnitude of charge for which the force is to be obtained in $\\mu$C, randomised in steps of 0.5$\\mu$C from 0.5 to 2.5.
", "templateType": "anything"}, "force1": {"name": "force1", "group": "Ungrouped variables", "definition": "{q1}*{q2}*(10^(-12))/(4*pi*{eps0}*{epsair}*(r1*(10^(-2)))^2)", "description": "Coulomb force on one charge due to another, in N, is given by $q_1q_2/4\\pi\\varepsilon r^2$.
", "templateType": "anything"}}, "variablesTest": {"condition": "", "maxRuns": 100}, "ungrouped_variables": ["eps0", "epsair", "epsmat", "force1", "force2", "g", "mass", "q1", "q2", "r1"], "variable_groups": [], "functions": {}, "preamble": {"js": "", "css": ""}, "parts": [{"type": "gapfill", "useCustomName": true, "customName": "Force between two point charges", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "Using Coulomb's Law, calculate the force on a {q1} $\\mu$C point charge due to the electric field from a {q2} $\\mu$C charge when they are separated by {r1} cm of air.
\n$|\\vec{F}| =$ [[0]] Newtons.
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\nThe force is reduced by [[0]] %
", "gaps": [{"type": "numberentry", "useCustomName": true, "customName": "percentage", "marks": "2", "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "minValue": "(force1-force2)/force1*99.5", "maxValue": "(force1-force2)/force1*100.5", "correctAnswerFraction": false, "allowFractions": false, "mustBeReduced": false, "mustBeReducedPC": 0, "precisionType": "sigfig", "precision": "4", "precisionPartialCredit": 0, "precisionMessage": "You have not given your answer to the correct precision.", "strictPrecision": true, "showPrecisionHint": false, "notationStyles": ["plain", "en", "si-en", "scientific"], "correctAnswerStyle": "plain"}], "sortAnswers": false}, {"type": "gapfill", "useCustomName": true, "customName": "Balancing gravitational force", "marks": 0, "scripts": {}, "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "unitTests": [], "showCorrectAnswer": true, "showFeedbackIcon": true, "variableReplacements": [], "variableReplacementStrategy": "originalfirst", "nextParts": [], "suggestGoingBack": false, "adaptiveMarkingPenalty": 0, "exploreObjective": null, "prompt": "If in this particular case the gravitational force at the surface of the Earth balances the electric force on the $\\var{q1}\\,\\mu\\text{C}$ point charge due to the $\\var{q2}\\,\\mu\\text{C}$ point charge, what is its mass? Take the accelleration due to gravity at the Earth's surface to be {g} N/kg.
\nThe mass balanced by the electric field force is [[0]] kg
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