1840 United States presidential election in Georgia
The 1840 United States presidential election in Georgia took place between October 30 and December 2, 1840, as part of the 1840 United States presidential election. Voters chose 11 representatives, or electors to the Electoral College, who voted for President and Vice President.
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| Elections in Georgia |
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Georgia voted for the Whig candidate, William Henry Harrison, over Democratic candidate Martin Van Buren. Harrison won Georgia by a margin of 11.56%.
Results
| United States presidential election in Georgia, 1840[1] | ||||||||
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| Party | Candidate | Running mate | Popular vote | Electoral vote | ||||
| Count | % | Count | % | |||||
| Whig | William Henry Harrison of Ohio | John Tyler of Virginia | 40,339 | 55.78% | 11 | 100.00% | ||
| Democratic | Martin Van Buren of New York | Richard M. Johnson of Kentucky | 31,983 | 44.22% | 0 | 0.00% | ||
| Total | 72,322 | 100.00% | 11 | 100.00% | ||||
References
- "1840 Presidential General Election Results - Georgia". U.S. Election Atlas. Retrieved 23 December 2013.
State and district results of the 1840 United States presidential election | ||
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